THE LIBRARY OF THE UNIVERSITY OF CALIFORNIA LOS ANGELES Tk RALPfl D. REED LIBRART DEPARTMENT OP GHKLOGT ijMivKKarrr ^ California UM ANGCLEB. CAUF, Digitized by tine Internet Arciiive in 2007 witii funding from IVIicrosoft Corporation littp://,www.arcliive.org/details/calculusOOmarciala CALCULUS MODERN MATHEMATICAL TEXTS EDITED BY Charles S. Slighter ELEMENTARY MATHEMATICAL ANALYSIS By Charles S. Slighter 490 pages, 5 x IM, Illustrated. MATHEMATICS FOR AGRICULTURAL STUDENTS By Henry C. Wolff 311 pages, 5 x 7J.i, Illustrated . CALCULUS By Herman W. March and Henry C. "Wolff 360 pages, 5 x TVi, Illustrated . MODERN MATHEMATICAL TEXTS Edited by Charles S. Slighter CALCULUS BY HERMAN W. MARCH, Ph. D. ASSOCIATE PKOFE880R OP MATHEMATICS UNIVEKSITY OF WISCONSIN AND HENRY C. WOLFF, Ph. D. PBOFESSOR OF MATHEMATICS DREXEIi INSTITUTE First Edition Sixth Impression McGRAW-HILL BOOK COMPANY, Inc. NEW YORK: 370 SEVENTH AVENUE LONDON: 6 & 8 BOUVERIE ST., E. C. 4 1917 Copyright, 1917, by the McGraw-Hill Book Company, Inc. THE MAPLE PRESS YORK PA Geolo^^ Library PREFACE One of the purposes of the elementary working courses in mathe- matics of the freshman and sophomore years is to exhibit the bond that unites the experimental sciences. "The bond of union among the physical sciences is the mathematical spirit and the mathematical method which pervade them." For this reason, the applications of mathematics, not to artificial problems, but to the more elementary of the classical problems of natural science, find a place in every working course in mathematics. This presents probably the most difficult task of the text-book writer, — namely, to make clear to the student that mathematics has to do with the laws of actual phenomena, without at the same time undertaking to teach technology, or attempting to build upon ideas which the student does not possess. It is easy enough to give examples of the application of the processes of mathematics to scientific prob- lems; it is more difficult to exhibit by these problems, how, in mathematics, the very language and methods of thought fit naturally into the expression and derivation of scientific laws and of natural concepts. It is in this spirit that the authors have endeavored to develop the fundamental processes of the calculus which play so important a part in the physical sciences; namely, to place the emphasis upon the mode of thought in the hope that, even though the student may forget the details of the subject, he will continue to apply these fundamental modes of thinking in his later scientific or technical career. It is with this purpose in mind that problems in geometry, physics, and mechanics have been freely used. The problems chosen will be readily comprehended by students ordinarily taking the first course in the calculus. A second purpose in an elementary working course in mathe- matics is to secure facility in using the rules of operation which must be applied in calculations. Of necessity large numbers of drill problems have been inserted to furnish practice in using the 104^284 vi PREFACE rules. It is hoped that the solution of these problems will be re- garded by teacher and student as a necessary part but not the vital part of the course. While the needs of technical students have been particularly in the minds of the authors, it is believed that the book is equally adapted to the needs of any other student pursuing a first course in calculus. The authors do not believe that the purposes of courses in elementary mathematics for technical students and for students of pure science differ materially. Either of these classes of students gains in mathematical power from the type of study that is often assumed to be fitted for the other class. In agreement with many others, the book is not divided into two parts. Differential Calculus and Integral Calculus. Integration with the determination of the constant of integration, and the definite integral as the limit of a sum, are given immediately fol- lowing the differentiation of algebraic functions and before the differentiation of the transcendental functions. With this arrange- ment many of the most important applications of the calculus occur early in the course and constantly recur. Further, with this arrangement, the student is enabled to pursue more advantageously courses in physics and mechanics simultaneously with the calculus. The attempt has been made to give infinitesimals their proper importance. In this connection Duhamel's Theorem is used as a valuable working principle, though the refinements of statement upon which a rigorous proof can be based have not been given. The subjects of center of gravity and moments of inertia have been treated somewhat more fully than is usual. They are par- ticularly valuable in emphasizing the concept of the definite integral as the limit of a sum and as a mode of calculating the mean value of a function. Sufficient solid analytic geometry is given to enable students without previous knowledge of this sub- ject to work the problems involving solids. In the last chapter simple types of differential equations are taken up. The book is designed for a course of four hours a week through- out the college year. But it is easy to adapt it to a three-hour course by suitable omissions. The authors are indebted to numerous current text-books for many of the exercises. To prevent distracting the student's at- PREFACE vii tention from the principles involved, exercises requiring compli- cated reductions have been avoided as far as possible. The book in a preliminary form has been used for two years with students in the College of Engineering of the University of Wisconsin. Many improvements have been suggested by our colleagues, Professor H. T. Burgess, Messrs. E. Taylor, T. C. Fry, J. A. Nyberg, and R. Keffer. Particular acknowledgment is due to the editor of this series. Professor C. S. Slichter, for suggestions as to the plan of the book and for suggestive criticism of the manu- script at all stages of its preparation. The authors will feel repaid if a little has been accomplished toward presenting the calculus in such a way that it will appeal to the average student rather as a means of studying scientific prob- lems than as a collection of proofs and formulas. Unfversity op Wisconsin, Herman W. March, November 6, 1916. Henry C, Wolff. CONTENTS Page Preface v Introduction 1 1. Constant. Variable. Function . . - 1 2. The Power Function 1 3. The Law of the Power Function 3 4. Polynomials. Algebraic Function 4 5. Transcendental Functions " . 6 6. Translation 6 7. Elongation or Contraction, or Orthographic Projection, of a Curve . 7 8. Shear 8 9. The Function a' 9 10. The Function sin x 9 11. The Functions p = a cos 0, p = b sin 0, and p = a cos d + 6 sin e 9 12. Fundamental Transformations of Functions 10 CHAPTER I Derivative 13. Increments 13 14. The Function y = x^ 16 15. Slope of the Tangent 17 16. Maxima and Minima 20 17. Derivative 21 18. Velocity of a Falling Body 22 19. Illustrations 23 20. Illustrations 23 CHAPTER II Limits 21. Definition 27 22. Notation 29 X CONTENTS Section Page 23. Infinitesimal 29 24. Theorems on Limits 30 25. The Indeterminate Form ^ 30 26. Continuous and Discontinuous Functions 31 CHAPTER III Thf Power Function 27. The Power Function 33 28. The Derivative of x» 33 29. The Derivative of ax" 35 30. Rate of Change of ax" 37 31. The Derivative of the Sum of a Function and a Constant . 39 32. The Derivative of au" 41 33. The Derivative of u", n a Positive Fraction 42 34. The Derivative of a Constant 43 35. The Derivative of the Sum of Two Functions 43 36. Differentiation of ImpHcit Functions 44 37. Anti-derivative. Integration 46 38. Acceleration 52 CHAPTER IV DiPFEKENTIATION OF AlGBBRATION FUNCTIONS 39. The Derivative of the Product of a Constant and a Variable. 56 40. The Derivative of the Product of Two Functions 57 41. The Derivative of the Quotient of Two Functions .... 58 42. The Derivative of w", n negative 59 43. Maximum and Minimum Values of a Function 00 44. Derivative of a Function of a Function 66 45. Inverse Functions 67 46. Parametric Equations 67 47. Lengths of Tangent, Normal, Subtangent, and Subnormal 68 CHAPTER V Second Derivative. Point of Inflection 48. Second Derivative. Concavity 70 49. Points of Inflection 71 CONTENTS xi CHAPTER VI Applications Section Page 50. Area Under Curve: Rectangular Coordinates 75 51. Work Done by a Variable Force 77 52. Parabolic Cable 82 53. Acceleration 83 54. The Path of a Projectile 85 CHAPTER VII Infinitesimals. Differentials. Definite Integrals 55. Infinitesimals 88 56. j;'!'.'^ • ■ ■ ■ 88 lim tan a lim tan a ^'- a = Q~^'(x^Q^vr~a. ^^ 58. ^^ l^^^ 89 a: = a 59. Order of Infinitesimals 90 60. Theorem 93 61. Differentials 95 62. Formulas for the Differentials of Functions 96 63. Differential of Length of Arc : Rectangular Coordinates . . 100 64. The Limit of 2;/(a;)Aa; 102 65. Definite Integral 103 66. Duhamel's Theorem 105 •67. Work Done by a Variable Force 107 68. Volume of a Solid of Revolution 109 69. Length of Arc: Rectangular Coordinates . Ill 70. Area of a Surface of Revolution 112 71. Element of Integration 113 72. Water Pressure 114 73. Airthmetic Mean 116 74. Mean Value of a Function 117 CHAPTER VIII Circular Functions. Inverse Circular Functions 75. The Derivative of sin m 122 76. The Derivative of cos u, tan m, cot u, sec m, and esc u . .124 xii CONTENTS Section Page 77. The Derivative of the Inverse Circular Functions .... 131 78. Velocity and Acceleration 136 79. Angular Velocity and Acceleration 139 80. Simple Harmonic Motion 140 81. The Simple Pendulum 141 CHAPTER IX Exponential and Logarithmic Functions 82. The Derivative of the Exponential and Logarithmic Func- tions 145 83. Logarithmic Differentiation 153 84. Compound Interest Law 155 85. Relative Rate of Increase 159 86. Hyperbolic Function 159 87. Inverse Hyperbolic Functions 160 88. The Catenary 161 CHAPTER X Maxima and Minima 89. The Maximum or Minimum of y = ax^ + fix + y . . . . 165 90. The Function a cos x + b sin x 165 91. The Function mx ± \^a^ - s" 168 92. Ma.xima and Minima by Limits of Curve 169 93. Maxima and Minima Determined by the Derivative . . . 169 94. Second-derivative Test for Maxima and Minima 172 95. Study of a Function by Means of its Derivatives 172 96. Applications of Maxima and Minima 174 CHAPTER XI Polar Coordinates 97. Direction of Curve in Polar Coordinates 178 98. Differential of Arc : Polar Coordinates 181 99. Area: Polar Coordinates 183 CONTENTS xiii CHAPTER XII Integration Section Page 100. Formulas 185 101. Integration of Expressions Containing ax^ + bx -\- c, by Completing the Square 190 102. Integrals Containing Fractional Powers of a: or of a + fca; . 192 103. Integrals of Powers of Trigonometric Functions 193 104. Integration of Expressions Containing \/a^ — x^,'\/a'^ + x^, '\/x^ — o*, by Trigonometric Substitution 196 105. Change of Limits of Integration 199 106. Integration by Parts 201 .07.Tho.„tegraU/e«si„n...,/.«c«„... .202 108. J sm3 xdx 204 109. Wallis' Formulas 204 ,^-, ., , ,. , Ta sin X + 6 cos X , 110. Integration of I — -. -. — j ax 209 J c sm x -\- a cos x 111. Partial Fractions 211 CHAPTER XIII Applications of the Process of Integration. Improper Integrals 112. Summary of Applications 208 113. Improper Integrals 223 114. Improper Integral: Infinite Limits 226 CHAPTER XIV Solid Geometry 115. Coordinate Axes. Coordinate Planes 228 116. The Distance Between Two Points 230 117. Direction Cosines of a Line 230 118. Angle Between Two Lines 231 119. The Normal Form of the Equation of a Plane 232 120. The Equation Ax + By + Cz == D 233 121. Intercept Form of the Equation of a Plane . 234 122. The Angle Between Two Planes 236 123. Parallel and Perpendicular Planes 236 xiv CONTENTS Section Page 124. The Distance of a Point From a Plane 237 125. Symmetrical Form of the Equation of a Line 238 126. Surface of Revolution 240 127. Quadric Surfaces 242 128. Cylindrical Surfaces 244 129. Partial Derivatives 246 130. Partial Derivatives of Higher Order 247 CHAPTER XV SuccEssrvE Integration. Center of Gravity. Moment OF Inertia 131. Introduction 250 132. Illustration of Double Integration 251 133. Area by Double Integration : Rectangular Coordinates . . . 253 134. Geometrical Meaning of the Definite Double Integral . . . 255 135. Area: Polar Coordinates 258 136. Volume of a Solid: Triple Integration 260 137. Center of Mass. Centroid 263 138. Centroid Independent of the Position of the Coordinate Planes 264 139. Center of Gravity . 264 140. Centroid of a Continuous Mass 265 141. Theorems of Pappus 274 142. Centroid: Polar Coordinates 276 143. Moment of Inertia . . . ' 277 144. Transfer of Axes 280 145. Moment of Inertia of an Area 281 146. Moment of Inertia: Polar Coordinates 283 147. Moment of Inertia of a Solid 284 CHAPTER XVI Curvature. Evolutes. Envelopes 148. Curvature 289 149. Curvature of a Circle 290 150. Circle of Curvature. Radius of Curvature. Center of Curvature 290 151. Formula for Curvature and Radius of Curvature: Rectan- gular Coordinates 290 CONTENTS XV Section Page 152. Curvature: Parametric Equations 292 153. Approximate Formula for the Curvature 293 154. Center of Curvature. Evolute 294 155. Envelopes 296 156. The Evolute as the Envelope of the Normals 300 157. Involutes 302 CHAPTER XVII Series. Taylor's and Maclaurin's Theorems. Indeter- minate Forms 158. Infinite Series 303 159. Rolle's Theorem 305 160. Law of the Mean 306 161. The Extended Law of the Mean 307 162. Taylor's Theorem with the Remainder 308 163. Taylor's and Maclaurin's Series 311 164. Second Proof for Taylor's and Maclaurin's Series .... 313 165. Tests for the Convergence of Series 315 166. Computation of Logarithms 320 167. Computation of tt 321 168. Relation Between the Exponential and Circular Functions 321 169. DeMoivre's Theorem 323 170. Indeterminate Forms 324 CHAPTER XVIII Total- Derivative. Exact Differential 171. The Total Derivative 328 172. Exact Differential 332 173. Exact Differential Equations 335 174. Envelopes .336 CHAPTER XIX Differential Equations 175. Differential Equations 338 176. General Solution. Particular Solution 338 177. Exact Differential Equation 339 178. Differential Equations: Variable Separable 339 xvi CONTENTS Section Page 179. Homogeneous Differential Equations 340 180. Linear Differential Equations of the First Order 342 181. Extended Form of the Linear Differential Equation . . . 344 182. Applications 345 183. Linear Differential Equations of Higher Order with Con- stant Coefficients and Second Member Zero 347 184. Au.xiliary Equation with Equal Roots 349 185. Auxiliary Equation with Complex Roots 350 186. Damped Harmonic Motion 351 Index 154 CALCULUS INTRODUCTION 1. Constant. Variable. Function. 1. A symbol of number or quantity, as a, to which a fixed value is assigned throughout the same problem or discussion is called a constant. 2. A symbol of number or quantity, as x, to which a succession of values is assigned in the same problem or discussion is called a variable. Example. The mass or weight of mercury in a thermometer is constant. The number that results from measuring this quantity (weight) is a constant. The volume of the mercury in the thermometer is variable. The number that results from measuring this quantity (volume) is a variable. 3. The variable y is said to be a function of the variable x if, when X is given, one or more values of y are determined. 4. X, the variable to which values are assigned at will is called the independent variable, or the argument of the function. 5. y, whose values are thereby determined, is called the de- pendent variable. 6. y is said to be a function of several variables u, v, w, • • • if, when u, v, w, • • • are given, one or more values of y are determined. 7. The variables u, v, w, ' • • , to which values are assigned at will are called the independent variables, or the arguments of the function. Functions of a single variable or argument are represented by symbols such as the following: f{x), F{x), 4>{x), yp{x). Func- tions of several arguments are represented by symbols such as f{u, v, w), F{u, v, w), 0(w, V, w). 2. The Power Function. $. The function x", where n is a constant, is called the power function. 1 CALCULUS (§2 If n is positive the function is said to be of the parabolic type, and the curve representing such a function is also said to be of the parabolic type. If w = 2, the curve, y = x"^, is a parabola. -^y 4-Li'' Fig. 1.— Curves for y = x", n = 1, 2, 3, and 4. If n is negative the function x" is said to be of the hyperbolic type, and the curve representing such a function is also said to be §3] INTRODUCTION of the hyperbolic type. If n = —1, the curve, y = x-^, is an equilateral hyperbola. In Figs. 1, 2, 3, and 4, curves representing ?/ = x" for different values of n are drawn. In Fig. 1, n has positive integral values; in Fig. 2, positive fractional values; in Fig. 3, negative integral values; and in Fig. 4, negative fractional values. The curves for StY" Fig. 2. — Curves for y = x", n = 5, 5, f, and |. y = x" all pass through the point (1, 1). They also pass through the point (0, 0) if n is positive. If n is negative, they do not pass through (0, 0). In the latter case the coordinate axes are asymptotes to the curves. 3. The Law of the Power Function. 9. In any power function, if X changes by a fixed multiple, y also changes by a fixed multiple. The same law can be stated as follows: CALCULUS m 10. In any power function, if x increases by a fixed percent, y also increases by a fixed percent. The preceding statements are also equivalent to the fol- lowing: 11. In any power function, if x runs over the terms of a geometrical progression, then y also runs over the terms of a geometrical progression. Fig. 3. — Curves for y = x", n = —1, —2, and —3. 4. Polynomials. Algebraic Function. 12. A polynomial in a; is a sum of a finite number of terms of the form ax", where a is a constant and n is a positive integer or zero. For example: ax^ + bx^ -^ ex -\- d. 13. A polynomial in x and y is a sum of a finite number of INTRODUCTION terms of the form ox'»i/'», where a is a constant and m and n are positive integers or zero. For example: axhj^ + ^xy^ + cx^ -{• dy -{• e. 1 i. Functions of a variable x which are expressed by means of a finite number of terms involving only constant integral and Y w=--|- Fig. 4. — Curves for y = x^, n ■= — |, — 5, — f, and — |. fractional powers of x and of polynomials in x are included in the class of functions known as algebraic Junctions^ of x. For example : (a) x\ (d) ^, + 1 + 1- (b) a;3 + (2a; 3)i. a;'' ■ X ie) X + 5 + (c) Vx^ + 4x + 7 + 4x + 5. (/) ■\/x — 7 3a;2 + 5a; + 7 x3 _ 3a; + 2 1 A function of x defined by the equation F(,x, y) = 0, where F{x, y) is a polynomial in X and y, is an algebraic function of x. For example, y = \/xs + 2 is an algebraio function of x. For by squaring and transposing we obtain 2/2 - x2 - 2 = 0, in which the first member is a polynomial in x and y. 6 CALCULUS f§5 15. An algebraic function is said to be rational if it can be expressed by means of only integral powers of x together with constants. Rational algebraic functions are divided into two classes: rational integral functions and rational fractional functions. 16. A rational integral function of x is a polynomial in x. 17. A rational fractional function is a quotient of two poly- nomials in X. It is usually desirable to reduce rational fractional functions of a: to a form in which the numerator is of lower degree than the denominator. This can always be done by performing long division. X + 3 2 Thus y = . 1 is equivalent to y = 1 -\ ^rjy, and Sx^ + 5x + 7 . . , ,, o , Ux + 1 5. Transcendental Functions. The circular (or trigonometric), the logarithmic, and the exponential functions are included in the class of functions known as transcendental^ functions. 6. Translation. If, in the equation of a curve fix, y) = 0, X is replaced by (x — a), the resulting equation, fix - a, y) = 0, represents the first curve translated parallel to the axis of a; a distance a; to the right if a is positive; to the left if a is negative. If y is replaced by (y — /3) the resulting equation, fix, 2/ - /3) = 0, represents the original curve translated parallel to the axis of y a distance j8; up if /3 is positive; down if /3 is negative. Thus V = (x + 3)^ — 4 is the parabola y = x- translated three units to the left and four units down. See Fig. 5. * All functions which are not algebraic functions as defined by the footnote on p. 6 are transcendental functions. §7] INTRODUCTION 7. Elongation or Contraction, or Orthographic Projection, of a X The substitution of - for x in the equation of any locus Locus multiplies all of the abscissas by a. X-f This transformation can be considered as the orthographic projection of a curve lying in one plane upon another plane, the two planes intersecting in the axis oiy. If a < 1 the second curve is the projection of the former curve upon a second plane through the F-axis and making an angle a, whose cosine is equal to a, with the first plane. If a > 1, the first curve is the projection of the second when the cosine of the angle between their planes . 1 IS -. CALCULUS y Similarly the substitution of - for y in the equation of a locus multiplies the ordinates by a. The interpretation from the standpoint of orthographic projection is evident from what has just been said. See Figs. 6 and 7. 8. Shear. The curve y = f{x) + mx is the curve y = f{x) sheared in the line y = mx in such a way that the y-intercepts remain unchanged. Every point on the curve y = f{x) to the right of the F-axis is moved up (down if m is negative) a distance proportional to its abscissa; and every point to the left of the Y- axis is moved down (up if m is negative) a distance proportional to its abscissa. The factor of proportionality is m. In general a curve is changed in shape by shearing it in a line. The parabola is an exception to this rule. Thus y = dx^ sheared in the line y = mx becomes y = ax^ + mx, or y = ^['' + 2a) -ia' This may also be considered as the result of translating the §11] INTRODUCTION 9 4M tfl original curve by the amounts — ^ and — j- in the x and ij direc- tions, respectively. Hence, by shearing, the parabola y = az^ is merely translated. 9. The Function a*. In Fig. 8 are given the graphs of ?/ = a*, for the values a = 1, 2, and 3. By reflecting these curves in the line y = X we have the corresponding curves for y = logo x. The exponential function y = a" has the property that if x is given a series of values in arithmetical progression the corre- sponding values of y are in geometrical progression. Fig. 8. — Curves for y = a*, a = 1, 2, and 3. 10. The Function sin x. The function y = sinx is repre- sented in Fig. 53. 11. The Functions p = a cos 9, p = 6 sin ^, and p = a cos 6 + b sin 6. The function p = a cos d is the circle OA, Fig. 9, and p = bsin d is the circle OB, Fig. 9. The function p = a cos 6 + 6 sin d can be put in the form p = R cos {8 — a), where 10 CALCULUS [§12 R = \/a^ + 6S and where cos « = n and sin a = „ • This function is represented by a circle, Fig. 9, passing through the pole, with diameter equal to R, and with the angle AOC equal to a. The maximum value of the function is R and the minimum value is -R. 12. Fimdamental Transformations of Functions. It is valuable to formulate the transformations of simple functions, that most commonly occur, in terms of the effect that these transformations have upon the graphs of the functions. The following list of theorems on loci contains useful facts concerning these transformations : Fig. 9. THEOREMS ON LOCI I. If X be replaced by {—x) in any equation containing x and y, the new graph is the reflection of the former graph in the F-axis. II. If y be replaced by {—y) in any equation containing x and y, the new graph is the reflection of the former graph in the X-axis. III. If X and y be interchanged in any equation containing x and y, the new graph is the reflection of the former graph in the line y = X. IV. Substituting / - ) for x in the equation of any locus multipUes all abscissas l)y a. §12] INTRODUCTION 11 V. Substituting L\ for y in the equation of any locus multiplies all ordinates of the curve by b. VI. If (x — a) be substituted for x throughout any equation, the locus is translated a distance a in the a;-direction. VII. If (y — b) be substituted for y in any equation, the locus is translated the distance b in the ^/-direction. VIII. The addition of the term mx to the right side of y = fi^) shears the locus y = f(x) in the line y = mx. IX. li {6 — a) he substituted for 9 throughout the polar equa- tion of any locus, the curve is rotated about the pole through the angle a. X. If the equation of any locus is given in rectangular coordi- nates, the curve is rotated through the positive angle a by the substitutions x cos a + y sin a for x and y cos a — x sin a for y. Exercises 1. Translate the curves (a) y = 2x2, (e) y = e", (t) y = -' (b) y = - 3X2, (_/■) y = ^3^ (y) y = _. (c) y = log X, (g) y = sin x, (/b) y = x^, 2 (d) y = e-^, (h) y = cos x, (l) y = x^, two units to the right; three units to the left; five units up; one unit down; two units to the left and one unit down. Sketch each curve in its original and translated position on a sheet of squared paper. 2. Shear each curve given in Exercise 1 in the line 2/ = ix; y = — \x', y = x; y = — x. Sketch each curve in its original and shearetl position. 3. Write the equation of each curve given in Exercise 1 when re- flected in the X-axis; in the F-axis; in the line y = x] in the line y = — X. Sketch each curve before and after reflection. 4. Rotate tlie ^curves 12 CALCULUS [§12 (a) p = a sin 9, (e) p = a(l + cos $), (b) p = a cos 6, (f) p = a(l — sin 6), (c) p = a cos ^ + 6 sin ^, (^) p = a(l + sin 0), (d) p = a (1 — cos 6), (h) p = ad, about the pole through an angle o' 7-' k' ''^j ~ o' *^ketch each curve in its original and rotated position. 6. Sketch the following pairs of curves on squared paper: (a) y = x^ and y = x^ -{- x. (6) y = x^ and y = (x — Sy + 2. (c) y = x^ and y = — a;^ — 2x. (d) 2/ = X* and y = a;* — 4a;' + Gx'^ — 4a;. (e) y = — 2x'^ and j/ = fx^. (f) y = x^ and 2/ = |x^. (?) 2/ = sin X and y = sin 2x. (h) y = sin x and y = 2 sin x. (i) y = cos X and y — sin ( „ — x j . 6. Rotate the following curves about the origin through the angle indicated. (o) x* - J/2 = a" through 45°. (6) x2 - y2 = a2 through -45°. (c) x2 - 7/2 = a2 through 90°. (d) a;2 - ?/ = a2 through -90°. (e) a;2 + 2/2 = a through a. if) y = mx^ through a. CHAPTER I DERIVATIVE In Elementary Analysis the student investigated the dependence of a function upon one or more variables with the help of algebra and geometry. He is now to study a very powerful method of investigating the behavior of functions, the method of the infinitesimal calculus, which was discovered by Newton and Leibnitz in the latter part of the 17th century. This method has made possible the great development of mathematical analysis and of its applications to problems in almost every field of science, particularly in engi- neering and physics. 13. Increments. Let us consider the following examples which illustrate the principles of the calculus: Example 1. A steel bar, subjected to a tension, will stretch, and the amount of stretching, or the elongation, will continue to increase as the intensity of the force applied increases, until rupture occurs. The elongation is a function of the applied force. In fact, if the force is not too great, so that the elastic limit is not exceeded, experiment has shown that the elongation is propor- tional to the applied force (Hooke's Law). If we denote the elongation by y and the force by x, the functional relation between them will be expressed by the simple equation y = kx, where A; is a constant. This relation is represented graphically by a straight line through the origin, Fig. 10. Suppose that after the bar has been stretched to a certain length, the force is changed. This change in the force produces a cor- responding change in the elongation, an increase if the force is increased, a decrease if the force is decreased. Evidently, from the law connecting the elongation and the force, this change in the 13 14 CALCULUS [§13 elongation is directly proportional to the change in the force. We shall call the change in the force x, the increment of the Jorce, or the increment of x, and shall denote it by the symbol A.x (read "in- crement of x" or "delta x"). The corresponding change in the elongation we call the increment of the elongation, or the increment of y, and denote it by Ly. In Fig. 10 let P be any point on the line y — kx. If x takes on ■an increment Ax, y takes on an increment Ay. We see that the Ay . ratio of these mcrements, i.e., the quotient -r- is entirely inde- pendent of the magnitude and sense of Ax and of the position of P on the line. Indeed this ratio is the slope of the line. Here the increment of y is everywhere k times the increment of x. Fig. 10. The relation between Ay and Ax can be shown without the use of the figure as follows: If x is given the increment Ax, y takes on an increment Ay so that y + Ay = k{x + Ax). On subtracting y = kx, Ay = kAx. Hence Ax ' a quantity independent of x and of Ax. Example 2. A train is moving along a straight track with a constant velocity, i.e., it passes over equal distances in equal inter- §13] DERIVATIVE 15 vals of time. Denoting by s the distance measured, say in miles, from a fixed point, and by t the time measured, say in hours, which has elapsed since the train passed this point, the functional relation between s and t is expressed by s = ct, where c is a constant denoting the velocity of the train. This function is represented graphically by a straight line. Fig, 11. If we take an increment of time At following an instant t and measure As the distance As passed over in this time, the quotient r-. repre- sents the velocity of the train, since we have assumed the velocity As of the train to be uniform. Furthermore, the quotient -r^ will oo >^ i 5 ^^^t AS ^ o Time ( t ) Fig. 11. be independent of the length of A^ and of the time t to which the increment was given. As is everywhere c times A^. This is evident from the graph. In these two examples the functions were both linear functions of the independent variable. We have seen in these cases (and clearly the same is true for any linear function, y = ax -\- h) that the ratio of A?/ to Ax is constant. Ay is everywhere equal to a constant times Aa;, no matter how large Ax is taken and no matter at what point (x, y) on the graph the ratio is computed. Example 3. Let us now take an example in which the func- tional relation is no longer a linear one. We shall find that the ratio of the increment of the function to the increment of the variable is no longer constant. Suppose that the train of Example As 2 is not moving with constant velocity. Then the quotient -r--. 16 CALCULUS [§M IS called the average velocity of the train during the interval of time At. Evidently this quotient will approximate more and more closely to a fixed value the smaller the interval of time At, is chosen. The limiting value As of the quotient -rr as At approaches zero is called the velocity at the time t. Let the curve of Fig. 12 repre- sent graphically the relation be- As — tween s and t. The ratio t-; cal- Tlme (t) Fia. 12. At culated at any point P on the curve is no longer constant as in Example 2, but varies with At and also with the position of the point P. the power function 14. The Fxinction y = x^. Consider y = x^. Let us find the ratio of Ay to Ax at a certain point of the curve, say (0.2, 0.04), for different values of Ax. The results are given in the adjoining table. We observe that as Ax is taken smaller Ay and smaller the ratio v^ approaches more and more closely a value in the vicinity of 0.4. Ay The value of -r— will now be calculated Ax for any point P, (x, y), on the curve y =x^. From this value, which is a function of X and Ax, the limiting value as Ax approaches zero will be found. The point P, Fig. 13, has the abscissa x. If we give to x an increment Ax, we have corresponding to the abscissa, x + Ax, the point Q on the curve. Its ordinate is y + Ay •= {x -\- Ax)2. Ay is equal to the difference between the ordinates of P and Q, or Ay— (x + Ax)^ — x^ = 2x Ax + (Ax)2. Ax Ay Ay Ax 0.4 0.32 0.8 0.2 0.12 0.6 0.1 0.05 0.5 0.05 0.0225 0.45 0.02 0.0084 0.42 0.01 0.0041 0.41 0.005 0.002025 0.405 0.002 0.000804 0.402 0.0010.0004010.401 \ 1 §15] Then DERIVATIVE 17 Ay Ax = 2x + Ax. As Ax approaches zero the first term jn the second member remains unchanged, while the second term approaches zero. It follows • • » ^y that the limiting value of v- as Ax approaches zero is 2x. This result is expressed by the equation I Fig. 13. Ay (read "limit of -r^ as Ax approaches zero"). When x = 0.2, limAy = 0.4. This is the limiting value which the ratio tabu- lated in the last column of the table above is approaching. When ^ = 3, l^o^x = 6. When x = |, H^.f^ = 1. Thus the formula just obtained enables us to calculate very easily, for any Ay value of X, the limit of -r- as Ax approaches zero. 16. Slope of the Tangent. — The curve of Fig. 14 is the graph of the function y = f(x). On this curve take the point P with 18 CALCULUS [§16 coordinates x and y, and a second point Q with coordinates X -\- Ax and y + Ay. Draw the secant PQ making tlie angle <^ with the X-axis and the tangent PT' making the angle t with the Ay X-axis. From the figure, -r- is the slope of PQ, or Ay Ax = tan (/>. As Ax is taken smaller and smaller the secant PQ revolves about the point P, approaching more and more closely as its limiting position the tangent PT', and tan <^ approaches tan t. (The Y / , T / J / P / n X O 4 Fig. 14. student will recall that the tangent to a curve at a point P is defined as the limiting position of the secant PQ as the point Q approaches P.) Hence hm Ay aj=oAx ^^^=0 ^ Hence ^x^qX ^^ equal to the slope of the tangent to the curve y = /(^) oi the point for which this limit is computed. In the case of the parabola, y = x^, the slope of the tangent at 2 §15] DERIVATIVE 19 the point (x, y) is 2x. This shows that the curve becomes steeper and steeper for larger positive and negative values of x and that at X = the slope is zero. In Fig. 15, let the X-axis be divided uniformly and let the F-axis be divided in such a way that distances measured from on a uniform scale are equal to the squares of the numbers affixed to the points of division. Draw lines parallel to the F-axis through equidistant points on the X-axis and lines par- allel to the X-axis through points on the F-axis whose affixed numbers on the non- uniform scale are equal to the numbers affixed to the points on the X-axis through which lines were drawn. On the cross section paper thus constructed, any point at the intersection of a hori- zontal and a vertical line bearing the same number is a point on the curve ij = x^ which would be constructed in the usual way by using the uniform scale on the F- axis as well as on the X-axis. Join the consecutive points thus located by straight lines. These lines are the diagonals of the rectangles on the cross section paper and they are secants of the parabola y = x let PR = ^x Y 2 1 / 8 Q J it p 1 R Y 1 3 / 4 1 / 2 1 y 4 _^,,„0^^ „ ~ I - ^ ' . . f - r ; [ Fig. 15. Let PQ be such a diagonal and RQ Ay Then RQ = Ay and ^ = ^. the slope of the secant PQ. The diagonals give an approximate idea of the slope of the curve. The construction shows why the slope increases so rapidly with x. 20 CALCULUS [§16 As more and more horizontal and vertical lines are inserted, the diagonals approach more and more nearly the direction of the tangent lines. The fact that the slope of the tangent to the parabola y = x^ is 2x furnishes an easy way of constructing the tangent at any point P {x, y). We have only to draw from P in the direction of the positive X-axis, a line PK of unit length, and from the ex- tremity of this line, a line KT parallel to the F-axis, whose length is twice the abscissa of P. The line joining PT" is the tangent to the parabola at P. When the abscissa is negative the line XT is to be drawn downward. 16. Maxima and Minima. The algebraic sign of ^^q ~ enables us to tell at once where the function y is increasing and where it is decreasing as x increases. For, if the slope is positive at a point, the function is increasing with x at that point and the greater the slope the greater the rate of increase. Similarly if the slope is negative, the function is decreasing as x increases. Hence the function y = x^ is a decreasing function when a;<0 and an increasing function when x>0, since the slope is equal to 2x. When X = the slope is zero and the tangent is parallel to the X-axis. Since the function is decreasing to the left of x = and increasing to the right of this line, it follows that the function decreases to the value zero when x = and then increases. This value zero is a minimum value of the function y = x"^. In general we define minimum and maximum values of a function as follows: Definition. Let y = f{x), where /(x) is any function of a single argument. If y decreases to a value m as x increases and then begins to increase, m is called a minimum value of the function. If y increases to the value M as x increases and then begins to decrease, M is callec a maximum value of the function. Thus in Fig. 16, if ABDFHI is the graph oi y = f(x), the func- tion increases to the value represented by the ordinate bB and then begins to decrease. bB is then a maximum value of the function. Similarly fF is another maximum value. dD and hH are minimum values of the function. In referring to the graph of a function, points corresponding to maximum and points corresponding to minimum values of the §17] DERIVATIVE 21 function will be called, respectively, the maximum and minimum points of the curve. Thus B and F, Fig. 16, are maximum points and D and H are minimum points of the curve. Thus, zero is a minimum value oiy == x^ or (0, 0) is a minimum point on the curve y = x^. It will be noticed that a maximum value, as here defined, is not necessarily the largest value of the function, nor is a minimum value the smallest value of the function. A maximum value may even be less than a minimum value. 17. Derivative. We see that the limit of the ratio of the incre- ment of the function to the increment of the independent variable as the latter increment approaches zero, is very useful in studying the behavior of the function. This limit is called the derivative of the function with respect to the variable. Hence the following definition: The derivative of a function of a single independent variable with respect to that variable is the limit of the ratio of the increment of the function to the increment of the variable as the latter increment approaches zero. The derivative of a function y with respect to a dy variable x is denoted by the symbol ^' This symbol will not be considered at present as representing the quotient of two quan- tities but as a symbol for a single quantity. Later it will be interpreted as a quotient. (See §G1.) It is read, "the derivative of y with respect to x." The process of finding the derivative is called differentiation. 22 CALCULUS [§18 18. Velocity of a Falling Body. As a further illustration of the application of the derivative let us attempt to find the velocity of a falling body at any instant. The law of motion has been experi- mentally determined to be s = yt\ where s is the distance through which the body falls from rest in time t. If s is measured in feet and t in seconds, the constant g is 32.2 feet per second per second, s is plotted as a func- tion of the time in Fig. 17. At any time t, let t take on an increment A^ s will take on an increment As, represented in the figure by the line RQ. Since s = \gt^, 1 •0 Ql o » a S AS P P/ X R Ac^t Time (t) Fig. 17. Hence s + As = ^git + Aty. As = hgit + Aty - hgt\ (1) or As = gtAt + hg{Aty. (2) This is the distance through which the body falls in the interval As At counted from the time t. The quotient -r- is the average velocity for the interval At. The velocity at t has been de- 1" As fined as ^™ ^^, i.e., as the derivative of s with respect to t. To find this limit divide (2) by At and obtain As At = gt + \gAt, the average velocity for the interval A^. From which lira ^ At = At = gt, or ds dt = gi, (3) (4) the velocity at t. Thus the velocity at the end of three seconds is 96.6 feet per second; at the end of four seconds, 128.8 feet per second. §20] DERIVATIVE 23 19. Illustration. As an example of the use of the derivative in studying the behavior of a function, let us consider the power function y = x^. y + Ay = (x + Axy, y + Ay = x^ -\- 3a;2(Ax) + 3x(Aj;)2 + {Axy, Ay = 3x2(Aa;) + 3x(Aa;)2 + (Ax)^, ^ = 3x2 + SxiAx) + (Ax)\ Then Hm Ay _ Ax=0 Ax~ T- = 3x2. ax For X = the derivative is equal to zero and consequently the tangent at (0, 0) is horizontal and coincides with the X-axis. For all other values of x the derivative is positive. This shows that the function is an increasing function for all these values of x. Where is the slope of the curve equal to 1? Equal to Vs? 20. Illustration. The solution of the following problem will further illustrate the use of the derivative. Find the dimensions of the gutter with the greatest possible carrying capacity ^ ^ p and with rectangular cross section, which can be made from strips of tin 30 inches wide by bending up the edges to form ^ the sides. See Fig. 18. Fig. 18. If the depth MR is determined, the width is also determined, since the sum of the three sides MR, PQ, and RQ is 30 inches. We seek to express the area of the cross section as a function of the depth. Denote the depth by X and the area by A. The width RQ is 30 — 2x. Hence A = (30 - 2x)x. In Fig. 19, A is plotted as a function of x. A first increases with x 24 CALCULUS [§20 and then decreases. The value of x for which A reaches its greatest value can be determined from a graph with a high degree of approximation. The derivative can be used to calculate accurately this value of x and this saves construction of an accurate graph. From to H, A is an increasing function. Its derivative is therefore positive for this part of the curve . From H to N the function A is decreasing. Its derivative is therefore negative for this part of the curve. At the point H the derivative changes sign, passing from positive values through zero to negative values. The abscissa of the point H can then be found by finding the Fig. 19. derivative of A with respect to x and determining where it changes sign. In this case the change of sign occurs where the derivative is equal to zero. We find by the method of increments ax Ax = 4(7.5 - x). = when x = 7.5. If a:<7.5, ~r~ is positive and A is an dx dx dA increasing function. If x>7.5, -7— is negative and A is a de- creasing function. This shows that A increases up to a certain value at a; = 7.5 and then begins to decrease. Hence the gutter §20] DERIVATIVE 25 will have the greatest cross section if its depth be made 7.5 inches. It is interesting to plot the derivative as a function of x on the same axes. See the dotted line, Fig. 19. The statements made concerning the derivative are verified in the graph. Exercises 1. Consider the function y = /(x) whose graph is given in Fig. 20. In what portions of the curve is the derivative positive? In what portions negative? Where is the derivative equal to zero? y =f(x) Fig. 20. 2. Find ,, U y = 3x*. For what values of x is the function in- creasing? For what values decreasing? At what point does the tangent line drawn to the curve representing the function, make an angle of 45° with the positive direction of the axis of x? Find the coordinates of the maximum or minimum points on the curve. 3. Answer questions asked in Exercise 2, if y = x'. 4. Answer questions asked in Exercise 2, ii y = x*. 5. Answer questions asked in Exercise 2, if y = x^. 6. Answer questions asked in Exercise 2, if y = x* — 2x + 3. 7. Answer questions asked in Exercise 2, ii y = -^ 2" + 2x — 6. 8. Find the derivative of \/x. Solution. Let y = \/x. Then y +Ay = y/x +Ax and Ay = y/x + Ax — Vx Ay _ \^x + Ax — \/x Ax ~ Ax 26 CALCULUS [§20 Rationalize the numerator : Ay 1 Ax Vx + Ax + Vx As Ax approaches zero the right-hand side of this equation approaches -7^- 7^- Then Vx + vx or lira A^ 1 2Vx 1 2\/x 9. Find the derivative of v/x - ^. lU. Find the derivative of \/x* - -4. CHAPTER II LIMITS In §17 the derivative was defined as the limit of a certain ratio. The word limit was used without giving its precise definition, as the reader was supposed to have a fair conception of the mean- ing of this term from previous courses in mathematics. How- ever, since the entire subject of the calculus is based on limit processes it is well to review the precise definition and to state certain theorems from the theory of limits. 21. Definition. // a variable changes by an unlimited number of steps in such a way that, after a sufficiently large number of steps, the numerical value of the difference between the variable and a constant becomes and remains, for all subsequent steps, less than any preassigned positive constant, however small, the variable is said to approach the constant as a limit, and the constant is called the limit of the variable. 1 K A Xi X2 Xs Xi B Fig. 21. Illustration 1. Let AB, Fig. 21, be a line two units in length, and let x be the distance from A to a point on this line. Suppose that X increases from by steps such that any value of x is greater than the preceding value by one-half of the difference between 2 and 2 — X this preceding value, i.e., by — ^ — Xi, Xi, Xz, Xi, • • - are the end points of the portions of the line representing the successive values of x. Then the lengths X\B = 1, x^B = ^, XzB = (^)^ XiB = (^)', • • • , XuB = (^)""^ are the successive differ- ences between the constant length 2 and the variable length x. This difference becomes and remains less than any preassigned length KB after a sufficient number of steps has been taken. 27 28 CALCULUS [§21 This is true however small the length KB is chosen. Therefore, by the definition of the limit of a variable, 2 is the limit of the variable x. Illustration 2. Consider the variable x"^ — 2. Give to x the 3(2" - 1) vnliiAQ ft ^ A iJ- A.1 A3. values u, 2 , 4 , a i 1 « , 32, 2" = 3[x-i]. which are chosen by starting with x = and giving to it successive increments which are one-half the difference between 3 and the 7 B 8 6 R 5 a 4 «3 ^ , 2 < ) 1 a ! 5 i \ Fig. 22. preceding value of x. The corresponding values of a;^ — 2 are given in the adjoining table. The corresponding points, excepting (0, - 2), are plotted in Fig. 22. From the table and the expression a;^ — 2 it is readily seen that the difference between 7 and the variable x^ — 2 becomes and remains less than any previously assigned quantity (such as KB, Fig. 22) after a sufficiently large number of steps. Therefore 7 is the limit of the vari- able a;2 — 2 as X approaches 3. Illustration 3, By giving x values nearer and nearer 2, the value of ?: becomes nu- ' x — 2 merically larger and larger. Indeed its numeri- cal value can be made greater than any preas- signed positive number however large by choosing x sufficiently X x'' -2 0.0 -2.00 1.5 0.25 2.25 3.06 2.62 4.89 2.81 5.91 2.91 6.45 2.95 6.72 2.98 6.86 2.99 6.93 §23] LIMITS 29 near 2. The variable does not approach a limit as x ap- proaches 2. Instead of doing so it increases without limit. // a variable changes by an unlimited number of steps in such a way that after a sufficiently large number of steps its numerical value becomes and remains, for all subsequent steps, greater than any preassigned positive number however large, the variable is said to become infinite. Illustration 3 of this section is an example of a variable which becomes infinite, or approaches infinity. 22. Notation. If in any limit process, the variable, say y, is a function of another variable, say x, the successive steps by which y changes are determined by those by which x changes. // y approaches a limit A, as x approaches a limit a, we say that the limit of y as X approaches a is A, and write lim ., _ x=a y = A. After what has just been said, the meaning of the two following expressions will be clear: lim y ^ ^ lim„ = ^o. In the second case a limit does not really exist. The form of expression is only a convenient way of saying that if x is taken sufficiently near a, the value of y can be made to become and remain greater in numerical value than any preassigned positive number however large. From the illustrations of the preceding section we have: 1. ""^ X = 2, where n is the number of steps taken. 2. ^i"^ z=3 Q lim X.3 (^^ - 2) "2= "'^ 23. Infinitesimal. In the particular case where the limit of a variable is zero, the variable is said to be an infinitesimal. An infinitesimal is a variable whose limit is zero. Thus Ay and Ax which were used in §§13, 14, and 15 are thought of as approaching 30 CALCULUS [§24 zero and are infinitesimals. Hence the derivative, §17, is defined as the limit of the quotient of two infinitesimals. Infinitesimals are of fundamental importance in the Calculus. Indeed the subject is often called the Infinitesimal Calculus. 24. Theorems on Limits. The following theorems concerning limits are stated without proof: Theorem I. 7/ two variables are always equal and if one approaches a limit, the other approaches the same limit. Theorem II. The limit of the sum of tivo variables, each of which approaches a limit, is equal to the sum of their limits. Theorem III. The limit of the difference of two variables, each of which approaches a limit, is equal to the difference of their limits. Theorem IV. The limit of the product of two variables, each of which approaches a limit, is equal to the product of their limits. Theorem V. The limit of the quotient of two variables, each of which approaches a limit, is equal to the quotient of their limits, provided the limit of the divisor is not zero. If the limit of the divisor is zero, the quotient of the limits in Theorem V has no meaning, since division by zero is an impossible operation. For, the quotient Q of two numbers A and B is defined as the number such that when it is multiplied by the divisor B, the product is the dividend A. Now if B is zero while A is not zero, there clearly is no such number. 25. The Indeterminate Form 5. If, in the quotient considered above, A is also zero, any number will satisfy the requirement, so that Q is not determined. One encounters exactly this diffi- a;2 _ 4 culty in seeking the value of ^ at a; = 2. Its value is not determined at this point but it is determined for all finite values of X different from 2. We define its value at a; = 2 as the limit of its value as x approaches 2. The student should construct a graph of this function. Usually we proceed as follows to find the desired limit. lim a;2 - 4 Hm x=2 a; _ 2 x=2 (x + 2) = 4. a;2 _ 4 The expression -_-^ is said to be indeterminate at x = 2, §26] LIMITS 31 since any one of an infinite number of values can be assigned to it. The determination of its limiting value as x approaches the value 2 is called the evaluation of the indeterminate form. Indeterminate forms of this and other types are frequently found in the Calculus. Thus V- is an indeterminate form for Ax 0. We have already seen in several cases how it can be evaluated. Exactly as in the example just given, we have sought the limit of the quotient as Ax approaches zero and not the quotient when Ax = 0, because the latter quotient has no meaning. Exercises 1. Determine the following limits, if they exist. (6) ^^gCOtX. , lim (a) ^cosx. (c) \"} sin -. Draw the curve for values of x between — tt and -f- jr. (d) ".'V,xsin"- 2. Evaluate the following indeterminate forms : (a) x2 -9 X -3 3. Find (&) X* + 6x2| symbol z=s=3 ' ' 3x3 + x^\x^O lim 3x2 _ jjj^ 4^_ jjj^ 4^2^ jj^^ 4x^ + 3 j;=oo a; » x=oo53.2> a;===co5^2> x= «= Qx^ ' Show that it is an indeterminate form. 26, Continuous and Discontinuous Functions, graphs of the following functions: 1 1. y = 2. y = x^. 3. y = 7x + Discuss the Draw the 1 4. 7/ = tan X. 5. y = sin x. 6- y = 3^. Hint. In 6, values in the vicinity of a; = should be carefully determined. Take a set of values of x approaching from the left and another set approaching it from the right. 7. 2/ = 3^ 3'' + 2 3' + 1 32 CALCULUS [§26 Study the vicinity of a; = 0. See 6. The functions 2, 5, and 7 are said to be continuous while 1, 3, 4, 6, and 8 are discontinuous. The meaning of these terms is obvious from the graphs that have been drawn. A precise definition follows : A function f{x) is said to be continuous atx = a if Ji^ fix) = /(a). In 1, 3, 4, 6, and 8, this condition is not satisfied at a; = 0, 0, ^, 0, and 0, respectively. In these examples the functions either become infinite for the values of x in question or approach different limits as the value of x is approached from larger or smaller values. A function f{x) is said to be continuous in an in- terval (c, d), i.e., the interval c ^ x ^ d, if it is continuous at every point in this interval. Thus the functions 2, 5, and 7 are continu- ous in any finite interval. The remaining functions are continu- ous in any interval not containing the points to which attention has been called. CHAPTER III THE POWER FUNCTION 27. In Chapter I the derivative of a function was found by what may be called the fundamental method, viz., by giving to the independent variable an increment, calculating the corresponding increment of the dependent variable, and finding the limit of the ratio of these increments as the increment of the independent variable approaches zero. This method is laborious and since it will be necessary to find derivatives in a large number of problems, rules will be established by means of which the derivatives of certain functions can be written down at once. The process of finding the derivative of a function is called differentiation. In this chapter we shall find the derivative of the power function, and study the function by means of this derivative. The graphs oi y = x", for various values of n, appear in Figs. 1, 2, 3, and 4. If n is positive, the curves go through the points (0, 0) and (1, 1), and are said to be of the parabolic type. In this case a;" is an increasing function of x in the first quadrant. If n is negative, the curves go through the point (1, 1) but do not go through the point (0, 0). They are asymptotic to both axes of coordinates. These curves are said to be of the hyperbolic type. In this case x^ is a decreasing function of x in the first quadrant. The law of the power function, as stated in §3, should be reviewed at this point. 28. Derivative of x°. Let y = x", where n is at first assumed to be a positive integer. y -{- Ay = (x + A.r)". y+ Ay = X- +nx--''Ax + "^^,"~-"^^x"-2 (Ax)^ + ■ +(Ax)". Ay = nx--^Ax + ^^^ a-"-^ {AxY+ ■ ■ +(Ax)". If a 33 34 CALCULUS [§28 Ay , , n{n - 1) Ax lim Ay or dy dx + .^ a;»-2 Aa;+ • • • +(Ax)»-i. = nx"" = nx"-^ (1) This proof holds when n is a positive integer. In §§33 and 42 it will be shown that the formula obtained holds for fractional and negative exponents. For the present we shall assume the formula true for these exponents. Illustrations. 1. ^ = 3.=. ax dx dx X* 3. ^' . W. dt dv ~ ""^ ~ 2\/y Exercises dy . . . Find -y- in each of the following fifteen exercises: 1. y = x^. 6. y = x. 11. y = x*. 2. y = x<. 7. y = x*. 12. y = x^. 3. y = x^ 8. y = x». 18. y = x"*. 3 _1 4. y = x^". 9. y = x^. 14. y = x ^. 6. y = -• 10. y = -^^ 15. y = -,- 16. Find the slope of each of the curves of Exercises 1-15 at the point (1, 1); also at the point whose abscissa is ^ and whose ordinate is positive. 17. By making use of the derivative, find for what values of x each function given in Exercises 1-15 is increasing; is decreasing. §29] THE POWER FUNCTION 35 18. How does the slope of j/ = a;" change with increasing x, if x is positive and if n is positive and less than 1? If n is positive and greater than 1 ? 19. Find where the slope of each curve given in Exercises 1-15 is equal to zero; equal to 1. 20. Find:riif: at (a) s = t\ (C) 8 = Vt' (e) s = ^yr^ (b) s=l,- (d) s = \- t^ CO . = ^-- 21. Find ^f if: at (a) y = IK (c) y = ^l (e) y = IK ib)y=l (d) y = 4- U)y=l- 29. The Derivative of ax°. In case the power function is writ- ten in the more general form ax", it is easy to see that the con- stant multiplier a will appear as a coefficient in all terms on the right-hand side of the equations in the proof in §28, and the derivative oi y = ax" is, therefore, f^ = nax'^-^, (1) or d(ax'') dx — nax"-^. (2) The proof of the formula is for positive integral values of n only, but as in §28 will be assumed for all commensurable exponents.^ Since ax"-^ is the given power function y = ax" divided by x, formula (1) may be written ' The relation of formula (2) to that of §28 is at once evident when it is recalled that the curve J/ = oi" can be thought of as obtained from the curve j/ = x" by stretching all ordinates in the ratio 1 : a. Then the slope of the tangent at a point of j/ = ox" is o times the slope of the tangent to y = i" at the corresponding point; i.e., d(ax") d(x'^) , = o = anx" . dx dx 36 CALCULUS [§29 The geometrical meaning of formula (3) is shown by Fig. 23. y The fraction - is the slope of the radius vector OP from the origin to the point P on the curve. Formula (3) states that the slope, at any point of the graph of the function, y = ax", is n times the slope of the radius vector OP. Thus, if n = 1, ?/ = ax" reduces to a straight line through the origin, and the line has the same slope as OP. If n = 2 the curve is the parabola y = ax^, and the slope of the curve is always twice that of OP. If n = —1 the r F ft 1 / o Fig. 23. curve is the rectangular hyperbola, y = curve is the negative of the slope of OP. Illustrations. d{7x^) 1. dx il dx = 14a;. d{5x-^) ^ d(x-^) and the slope of the dx = 5 dx = -lOx-3 = - 10 3. -^-iT-- = 6 ^TT- = 9^2 dt 4. If 2/ = 5x', dt dy dx §30] THE POWER FUNCTION 37 5. If V = ~2' ^~ = -2 x^ dx X Exercises Find -T~ in each of the following fifteen exercises . 1. y = 4x3. 6. 2/ = 2xi 11. y = -3x«. 2. J/ = 3\/x. 7. 2/ = 4x^. 12. ?/ = x. 3. y = 5x*. 8. y = fx"". 13. y = -x. 4. y = 3x. 9. 2/ = lOv^i. 14. 2/ = -3\/x. 2 4 4 6. 2/ = -,' 10. 2/ = --2- 15. 2/ = --3- 16. Find -yj in each of the following: (a) s == 2<2. (6) s = SVi". (c) s = -^tK civ 17. Find -jt in each of the following: (a) 2/ = 4v^. {b) y = - At. (c) ?/ = -Si^. 18. Find the slope of each of the curves given in Exercises 1-15, at the point whose abscissa is 1; at the point whose abscissa is |. 19. For what values of x is each function given in Exercises 1-15 increasing? Decreasing? Where, if at all, is the slope of each of these curves zero? 2 20. Draw the curves y = \x^, y = ~, y = x^, y = -y/x] and draw tangent lines to them at the points for which the abscissas are 1, 2, 3, and 4. Make a table showing the slope of the radius vector and the tangent line for each of these points. 30. Rate of Change of ax°. Let y = ax", where a; is a function of the time t. Since a: is a function of t,y is a, function of t. For example, y = Sx^, where x = t — 1. Let Ax and Ay be the increments of x and y, respectively, corre- Aw spending to the increment A^ of t. -rr is the average rate of dv change of y during the interval AL .r is the rate of change of y at the instant t. At any time, t y = ax". 38 CALCULUS [§30 At the time t + Af, y -\- Ay = a{x + Ax)". y -\-Ay= alx-+nx"-'Ax + Z ^ x»-''(Ax)H h (Ax)"l • Ay = aLx"-' + "^^^^ a;»-2Aa; + • • • + (A.t)"-i1 Ax. Aw r , > W(W "~ 1) OA ■ I /A N iTAX ^^ = a wx»-i + ^2"^ x"-2Ax + • • ■ + (Ax)»-i ^• As Af approaches zero, the expression within the brackets ap- proaches nx""^, and irr approaches -jr* Hence dy dx dt -°^^""'dr or . dt "*^ dt The rate of change of the function ax" is expressed in terms of x dx and of -TT, the rate of change of x. If then the rate of change of X for a given value of x is known, the rate of change of the func- tion for that value of x can be calculated. Illustration 1. The side of a square is increasing at the uni- form rate of 0.2 inch per second. Find the rate at which the area is increasing when the side is 10 inches long. Let X be the length of the side, and y the area of the square. dx dv Then ^rr = 0.2 and ^r is the rate of increase of the area. To find dt dt this rate of increase, differentiate the function y = x^. ^ -2x— • dt ^^ dt Since dx dt = 0-2. dt "-^ §31] THE POWER FUNCTION 39 dv When X = 10, -rr = 4. The area is increasing at the rate of 4 dv square inches per second. When x = 13, -tt = 5.2, the rate of change of the area at this instant. Illustration 2. A spherical soap bubble is being inflated at the rate of 0.2 cubic inch per second. Find the rate at which the radius is increasing when it is 1.5 inches long. dV Let r be the radius, and V the volume of the bubble, -rr = 0.2 at dv and ^7» the rate of increase of the radius, is to be found. V = AirrK ^ = ^^^^41- From which dr 1 dV dt ^ 47rr2 dt Since dV TT- = 0.2, and r = 1.5, dr 1 tit ~ on-n K \2 ~ 0.0071 inch per second. Exercises 1. Find the rate at which the surface of the soap bubble of Illustra- tion 2 is increasing when r = 1.5 inches. 2. If each side of an equilateral triangle is increasing at the rate of 0.3 inch per minute, at what rate is the area of the triangle increas- ing when the side is 6 inches long? 3. Water is flowing at a uniform rate of 10 cubic inches per minute into a right circular cone whose semi-vertical angle is 45°, whose apex is down, and whose axis is vertical. At what rate is the surface of the water in the cone rising, and at what rate is the area of this surface increasing when the water in the cone is 25 inches deep? 31. The Derivative of the Sum of a Function and a Constant. Sketch, on the same set of axes, the graphs of the functions: 40 CALCULUS [§31 y = x^; y = x^ — 5; y = x^ -{- 3; y = x^ -\- 10. Find -v- for each dy of the functions. Find t-, if y = x"^ -{- C, where C is any constant. Sketch a graph of any function y = f{x), and on the same set of axes, graphs oi y = fix) -\- C for several values of the constant dy C. What relation exists between t- for the different functions dx corresponding to the same value of cc? From these illustrations it is clear that the derivatives of all Junctions which differ only by an additive constant are the same. The reason for this is geometrically evident. For, the addition of a constant to a function has the effect of merely translating the graph of the function parallel to the F-axis. The slope corre- sponding to any given abscissa is clearly not changed by this translation. Hence, d[f(x)+C] d[f^x)l dx dx In particular d[axn-|-C] d[ax'i] _ = _- — = a dx dx Thus, if _ , . „ dy d{5x^) rll— 1 (1) (2) dx dx Exercises 15x2 1. Prove formula (2) above by the increment process. Differentiate : 2. y = 3x^ + 2. 11. 7/ = - 4x^ + 6. 3. y = 5\/x + 4. 12. y = - 3x* + 2. 4. y = 2x3 _ 3. 13. y = Tx^ - 3. 3 5. 2/ = — 5 + 5. 14. ?/ = 4x' + 5. 6. 2/ = 2t* + 7. 15. 2/ = ^ + 2. 7. s = 16<2 + 5. 16. 2/ = - ix« + 3. 8. « = 2y/t^ + 6. 17. 2/ = ix* + 2. 9. s = ^3 - 4. 18. 2/ = ?x* - 5. 10. X = 4<3 - 2. §32] THE POWER FUNCTION 41 32. The Derivative of au°. li y = aw, where w is a function of X and n is a positive integer, the student will prove, as in §30, that dy .du dx dx or d(au'') „ .du dx dx Illustrations. = 30x(x2 + 3)2. 2 d[2(x'' + 4)^+10] _rf[2(x2 + 4)^]_,^ ,(J(x2 + 4) dx dx dx = 2-4(a;2 + 4)3 2.r = 16x(x2 + 4)'. 3. If y = (2x2 + 1)2, find ^. dy _ cj(2x 2+l)2 _ , 1^ ^(2x2+ 1) •2'2x 4. If 77 = (x2 + l)t. = 2(2x2 + i).2-2x ^ = 8x(2x2 + 1) ^j- dy _ 3 dx ~ " (a:2+l)^ 2x = 3x(x2 + 1)5, and dy 3 (X2+1)^ dt = 3x(x2 + 1)^ dx dt' Exercises Fi nd^: ax y = (4x2 y = 5(2x y = 2(3 y = \/x y = (5- -2)3. 2 - 5)3. - 4X2)3 - X2)3. 6. y 7. y 8. y 9. y 10. y 1. 2. 3. 4. 6. = V9 = (3x2 = a/x^ = V2a -X2. + 7)2. -5. ;2 + 3 ;2 + l. 42 CALCULUS [§33 11- y = /^ ' • 15. y = VSx - 2 A^x2 + 1 7 "i 12. 2/ = -7===-. 16. y = 2 13. y = V(x« + 4). 17. y = y- 14. 2/ = V(x + 1). 18. 2/ = (a:2 + 4)2 3 (5 -x»)2 33. The Derivative of u*^, n a Positive Fraction. We are now in a position to prove that the rule for the derivative of m" holds when n is a positive fraction of the form -' where p and q are integers. Let p y = w. Raise each member to the power q: ?/« = UP. Since m is a function of x,yisa function of x. Hence each member is a function of x raised to a positive integral power. Then each member can be differentiated by the rule of §32 which was proved for positive integral exponents. We find , dy ^y-' Tx = = pu -1 '^'^. dx From which dy V tip-i 'du dx q y-^ dx Substitute for y in the second member and obtain dy P wp-^ du p up- -1 du dx ~ 9 u' dx q up- p dx Then dy ^1 dx q u du dx §35] THE POWER FUNCTION 43 and the rule is proved that dw , du ax dx where n is a positive fraction whose numerator and denominator are integers. This rule has already been used in the solution of numerous exercises. 34. The Derivative of a Constant. Let y = c, where c is a constant. Corresponding to any Ax, Ay = 0, and consequently Ax "' lim -^ = 0. and Ax = Ax or ^ = 0. dx The derivative of a constant is zero. Interpret this result geometrically. 35. The Derivative of the Sum of Two Functions. Let y = u + v, where u and v are functions of x. Let Au, Av, and Ay be the incre- ments of u, V, and y, respectively, corresponding to the increment Ax. y -\- Ay = u -{- Au + V -jr Av Ay = Au -\- Av Ay _ Au Av Ax ~ Ax Ax dy du . dv dx dx dx or d(u4-v) _ du dv dx ~ dx dx* The derivative of the sum of two functions is equoiL to the sum of their derivatives. 44 CALCULUS [§36 The student will observe that the proof given can be extended to the sum of three, four, or any finite number of functions. Illustrations. I t^(63^ + 15a;') ^ d(6x) rf(15x') _ g , 30^ dz dx dx ^ d{2V x + 3x' + 4) d{2V~x) , d(3x^) , d(4) 1 2. = h. ' = — 7= + ox. rfx dz dx dx y/x d(t^ + 2<' + 3) 3. f; = 2< + 6«2 Exercises Differentiate the following functions with respect to x, also with respect to t: 1. 3x* - 2x^ + 6. 10. ax2 + bx + c. 2. 5x» - 7x2 -2x - 10. 11. y = v^a;2 + 4^; _ 5. 3. gx* - ix' + X - 7. 12. 2/ = 1 4. x2 + 2x 3 2 y/x^ - 5x + 7 13. 2/ = V3x2 _ 2x + 5. * \7x ~ ^" ^ 14. y = Ve - 3x - x\ 6. 3x7 _ 63.8 + 9. 16. s = Vm + ^2< - 3 ?_2 1 .. -. — X ^ 16. 7/ = x" - 7x - 6 8. - \x^ -\- \x^ - X + 2. _ / II 17. y = V a;^ - 5x + 4. 9. x~5 4- a;~5. 18_ y _ (3_j.2 _|. 2x + 2)3. 36. Differentiation of Implicit Functions. The derivative of one variable with respect to another can be found from an equa- tion connecting the variables without solving the equation for either variable. For, if the variables are x and t/, j/ is a function of x, even though its explicit form may not be known, and the usual rules for finding the derivative of functions can be applied to each member of the equation. The following example will illustrate the process. dy Illustration. Letx' + 2/^ = O"- Find -? • The left-hand member of the given equation is the sum of §36] THE POWER FUNCTION 45 two functions of x, since y is a function of x. Further, the deriva- tive of the left-hand member is equal to the derivative of the right-hand member. The derivative of the latter is in this case zero, since the right-hand member is constant. On differentiat- x+ y=^o, Fig. 24. ing the left-hand member as the sum of two functions, we obtain Solving for -t-> 2x + 2y% = 0. dx When the derivative is found by differentiating each member of an equation in the implicit form, as in the foregoing illustration, the operation is called implicit differentiation. 46 CALCULUS [537 Exercises 1. Draw the circle x^ + y'' = o* and show geometrically that the slope of the tangent at the point (x, y) is dy 2. Solve the equation of Exercise 1 for y and find -j- • dv From the following equations find -r by implicit differentiation: 3. 3x« + 4y* = 12. 4. a;^ — ?/* = a^. ^' ^ + ^ = !• (^o ^'^^ clear of fractions.) If y is an implicit function of x expressed by an equation of the form x" + y = a", (1) differentiation gives or g=-[i]""' <^> The equation (1) includes a number of important special cases. The graphs corresponding to the following values of n are shown in Fig. 24. For n = h x' + y* = o^, a parabola, 2 1 2 . ■ n — I, x^ -\- y^ = a^, an important hypocycloid, n = l, X + y =a, a straight line, n = 2, x"^ -\- y"^ = o^, a circle. The graph of (1) passes through the points (0, a) and (a, 0) if n is positive. 12? 6. x» + J/* = o^. Ill 7. x* + 2/2 = o2. 8. x' + y' = a\ 3 3 3 9. x^ + y^ = a^. 37. Anti-derivatives. Integration. Let it be required to find the equation of a curve whose slope at any point is equal to twice the abscissa of that point. §37] THE POWER FUNCTION 47 dy This means that at every point of the curve -, = 2x. We seek then a function whose derivative is 2x. y = ^^ is such a function. But y = x^ -\- C, where C is a constant, is also a function having the same derivative. Hence there is an infinite number of functions whose derivatives are all equal to 2x. The problem as proposed has then an infinite number of solutions, viz., the system of parabolas y = x^ + C, corresponding to the infinitely many values of C. If now we add to the statement of the problem the requirement that the curve shall pass through a given point, say (1, 2), it is geometrically evident that but one of the curves y = x^ + C will pass through the point. In other words there is but one value of C for which the latter requirement is satisfied. This value is de- termined by substituting the coordinates of the point in the equation y = x^ -{- C, since they must satisfy this equation for some value of C, if the problem has a solution. On making the substitution we have 2 = 1 + C, from which (7=1. Hence y = x"^ -\- 1 is the equation of the curve whose slope at any point is equal to twice the abscissa of the point and which passes through the point (1, 2). The nature of the problem which has just been solved can be further explained by the following geometrical solution. Draw, Fig. 25, at the vertices of each small square on a sheet of co- ordinate paper on which a set of axes has been chosen, short lines whose slopes are equal to two times the abscissas of the respective vertices. A curve is to be drawn which at each of its points is tangent to a line such as those which have been drawn. Now it is impossible in the figure to draw lines through every point in the plane, but if the points through which the lines are drawn are sufficiently thick, the lines will serve to indicate the direction which the curve takes at nearby points. The lines may be regarded as pointers indicating the stream lines in flowing water. Then a point tracing the curve would move as a small cork would in water having the stream lines indicated by the figure. Thus, to get the curve that goes through (1, 2), start from this point and, guided by the direction lines, sketch in as accurately 48 CALCULUS I§37 as possible the curve to the right of this point. Do the same thing to the left, noting that here it is necessary to go against the stream lines instead of with them. In Fig. 25 it should be noted that all lines through points having the same abscissa are parallel. This fact is of great ^ ^ ^ ^ ^ \ ^ M ^ ^ H ^ k k h \ \ ^ k k k \^\^ /MM /^ ^ M M Fig. 25. assistance in drawing. The squares on the coordinate paper can be used to advantage in drawing lines when the slope is known. If the derivative which was given had been any other function of X, a geometrical solution could have been obtained by the same method. - The foregoing illustration introduces a new type of problem, §37] THE POWER FUNCTION 49 viz., that of finding a function whose derivative is given. A function whose derivative is equal to a given function is called an anti-derivative, or integral, of the given function. From the illustration it is clear that any given function which has one anti- derivative, has an infinite number of anti-derivatives which differ from each other only by an additive constant. This latter fact is indicated in obtaining the anti-derivative of a given function by writing down the variable part of the anti-derivative and adding to it a c onstant C which is undetermined or ' ' arbitrary. ' ' In a given application this constant will be determined by supplementary conditions as in the illustration at the beginning of this section. The process of finding the anti-derivative of a given function is called integration. y = x^ + C. y = lx' + C. y — x^ ■{- x^ -\- C. y = x^ + x"^ -\- Ix -\- C. 2/ = 3- + 2- + 7a; + C. If in Illustration 1 the curve is to pass through the point (3, -2) we must have -2 = 3' + C, or C = -29. Hence the equation of the curve is y = x^ — 29. Exercises Illustrations. 1. jfdy dx ~ 3x2 2. If — - dx x\ 3. -ffdy 3x2 + 2x, 4. If — - dx 3x2 + 2x -h 7, 5. Tf ^ - dx x^ + x + r. Integrate the following ten functions: 1- 1 = 3,. ^- dt ~ dx (3x2 + 2x + 6) ^ .. g = 4... ^ dy _ '' dt (ax + b) ^-. ''•|--4r ^- dx~ 3x2 _ 2x2 _|. 7. «• f = 3..f . Q '^y ^' dx- lOx-2 + 2x-3 - X 6. ^ = 3x2 + 2x - 6. 10. f~ = dx x^ + x'\ 60 CALCULUS l§37 Illustrations. 6. ^ = 3(x2 + 2)2 2x. The right-hand side is in the form, nw-^ ^, where n is 3, and w du -, is (x'^ + 2). Since the integral of nw"-i ^ is -u" + C, 2/ = (x^ + 2)» + C. 7. ^ = (x2-5)'2x = H4(x2-5)'2x]. 2/ = i(x2 - 5)^ + C. 8. ^ = x(x2- l)'' = -iM6(x2- l)''2x]. y = Mx^ - 1)« + C. y = _ -j^(3 - x')« + (7. 10. I = (.' - 2x + 3)-'(x - l);j^ = - il - 2(x^ - 2x + 3)-^(2x - 2)^J. y " 4(x2 - 2x + 3)2 + ^* Exercises Integrate : 11. ^ = a;V^^"^=l- Ans. y = K^^ - 1)^ + C. 12. ^ = (2x3 4. 3x2) 3 (x^ _|_ x). Ans. y = i(2x3 4. 3x2)1 + c, ax 13. ^ = (x + 1)*. ^ns- y = f (x + 1)^ + C. dx 14. I = (2- XT X. 16. ^^ MX- 3)2 X. 16.| = xVr^'. 17.g = x2+3x. 18.^^ = (x2 + 7)3x. §37] THE POWER FUNCTION 51 20. f^ = (x^ + 2x + iy(x + 1). 22. dy dx ~ X 24 dy dx X2 V5 -X* - (4 - x")* 25. dy dx ~ (3x2 + 2) ' X. Ans. y = A (3x2 + 2y + c. 26. dy dt ~ (2- - 3x2) . dx '"'dt- 27. dy ^ dx (2x + 1)^ Ans. y = K2x + 1] 3 + C. 28. dy dt ~ (3x -2y dx dt' 29.^f = (3-4x)2^. 30. ^ = V^TT- ^ns. 2/ = f (x + 1)' + C. 31. ^ = >/2F+^. 40. ^ = \/(2x+3)». 33. -2 = Vir=^. 42. ^ = Vx^ + 3x - 7 (x^ + 1). 3*- 5x = vTTT' "• ^' ^ ^" "^ 4)^/iM^8^+^. ^^- ^^ = virr2 '*• ^' = ^^ - ^)^^^^^+^- ^^ dy 1 .- dy 1 — 5x 36. T^ = , 46. " dx V3 - 5x * dx -y/e 4- 4a; - lOx* — dw / .^ dy 3x — 2 37. / = X V4x2 _ 5. 46. *' dx * ' dx V3x2 _ 4a; _|_ 5 88. I - .V9^'. «. I - V4^. 89. f . '^ 48. f^ - x(2 - x>).. dx .^^4 — 3x2 dx 49. Find the equation of the curve whose slope at any point is equal to the square of the abscissa of that point and which passes through the point (2, 3). 60. Find the equation of the curve whose slope at any point is equal 52 CALCULUS I §38 to the square root of the abscissa of that point and which passes through the point (2, 4). 51. Find the equation of the curve whose slope at any point is equal to the negative reciprocal of the square of the abscissa of that point and which passes through the point (1, 1). 38. Acceleration. The velocity of a body moving in a straight line may be either uniform or it may vary from instant to instant. In the latter case its motion is said to be accelerated, and this applies both to the case where there is an increase in velocity and the case where there is a decrease in velocity. Thus it is a fact of common knowledge that the velocity of a body falling to the ground from a height increases with the dis- tance through which the body has fallen, or with the time since the body started to fall. The time rate of change of the velocity of a moving body is an important concept in mechanics and physics. If s denotes the distance passed over in time t, the velocity has been defined as the rate of change of s with respect to t. The notion of the velocity at a given instant was derived from that of the average velocity for an interval A^ The average velocity was obtained by dividing the change in s, As, in a time A^ by At {i.e., by dividing the distance passed over in time At by At). The limiting value of this quotient as At approaches zero was defined as the velocity at the beginning of the interval A^. In the same way if the velocity, r, changes by an amount Av in the time A^ counted from a certain time t, the average rate of Av change of v for this interval is ^- It is the average linear accelerch tion^ for this interval. The acceleration at the time t is defined as the limit of the average acceleration as At approaches zero. It is then ^' The acceleration is the time rate of change of velocity. In the case of a falling body it is known experimentally that for bodies falUng from heights that are not too great, the velocity changes uniformly, due to the action of the force of gravity, i.e., • We suppose here that the body is moving in a straight line. If the path is curved, it will be seen later that the total acceleration is to be thought of as the resultant of two components, one of which produces a change in the direction of the velocity and the other a change in the magnitude of the velocity. §38] THE POWER FUNCTION 63 the time rate of change of the velocity is a constant. This con- stant is called the acceleration due to gravity and is usually denoted by g. In F.P.S. (foot-pound-second) units it is equal to 32.2 feet per second per second. That the unit of acceleration is 1 foot per second per second is explained by the fact that accelera- tion is the change per Second of a velocity of a certain number of feet per second. The differential equation of motion of the falling body can be written = 9- (1) dt From which on integrating, V = gt + C. If it is given that the body starts falling from rest, we have as the condition for determin- ing C, that V = when t = 0. Equation (2) shows that C must be equal to zero. Then, V = gt. (3) (2) Time U) Fig. 27. The graph, Fig. 27, o( v = gt is a, straight line whose slope is g. If the body had had an initial speed of Vq feet per second, i.e., if it had been projected downward instead of being dropped, the constant C would have been determined from the condition that 54 CALCTJLUS [§38 V = Vo when t = 0. It follows from (2) that C = vo, and the equation for v would have been V = gt + Vq. (4) The graph of this function is shown in Fig. 26. It is again a straight line but it cuts the F-axis at the point (0, Vo). The foregoing discussion evidently applies equally well to any uniformly accelerated motion, i.e., to any motion where the rate of change of the velocity is constant. In all such cases the graph of r as a function of the time is a straight line. Since v = -n' equation (4) gives ds di = 9i + f 0. Integrating, s = yt^ + vot + Ci. (5) If t is measured from the instant the body begins to move and s from the position of the body at that instant, s = when t = 0. From this condition d = 0. Then the distance of the body from its initial position is given by s = igf^ + Vot. If a body is thrown vertically upward, it is convenient to count distances measured upward, and upward velocities, as positive. Then, since the acceleration due to gravity diminishes v, equation (1) becomes di = -^- (^ ^ The formulas (4) and (5) then become v^ -gt + vo (4') s = - hgt' + Vot. (5') If a body falls from rest it is easy to express the speed as a function of the distance traversed. In this case, 2^0 = 0. Then (4) and (5) become, V = gt (4") i §38] THE POWER FUNCTION 55 Elimination of t between these equations gives: V = VWs- (6) Exercises 1. If a body falls from rest how far will it fall in 10 seconds? 2. If a body is thrown vertically downward with a velocity of 10 feet per second, how far will it have moved by the end of 10 seconds? What will its velocity be? 3. If a body is thrown vertically upward with a velocity of 64.4 feet per second, what will the velocity be at the end of 10 seconds? What will be the position of the body? How far will it have moved? 4. Find the laws of motion if the acceleration is equal to 2t and if (1°) s = and v = when t = 0; (2°) s = 3 and v = --2 when t = 0. 6. If the acceleration is proportional to the time and ii v = vo and 8 = So when i = 0, show that 8 = -^ + Vol + So. CHAPTER IV DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 39. The Derivative of the Product of a Constant and a Variable. Let y = cu, where c is a constant and u and y are functions of x. Let Am and Ay be the increments of u and y, respectively, corresponding to the increment Ax. Then 2/ + A?/ = c{u + Am) A?/ = cAm Ay Ax Am Ax dy Ix ~ du c v-» dx d(cu) dx ~ du C J* dx or The derivative of the product of a constant and a function is equal to the constant times the derivative of the function. Illustrations. d{3x^) d(x2) 1. — J = O —j — ■• = DX. ax dx ,m__m,_,^j^._,,^_,. 3. d[-f(x 2-5)2] dx ,d{x^ ^-5)^ -i d{x^ - dx = 3 2 dx 2-5)- X A) 3 (X2- 50 -5)2 §40] ALGEBRAIC FUNCTIONS 57 40. The Derivative of the Product of Two Functions. Let y = uv, where u and v are functions of x. y -\-Ay = (u + Au)(v + Aw) y + Ay = uv -\- nAv + vAu + AuAv Ay = uAv + vAu + AuAv Ay 4!i_i_ 4!f_i_A — Ax ~ Ax Ax Ax Since Aw approaches zero as Ax approaches zero, dy _ dv du dx dx dx or d(uv) dv , du ... ^^dr = ^dx + ^dx' ^^^ The derivative of the product of two functions is equal to the first times the derivative of the second plus the second times the derivative of the first. Illustraiions. - ^J^i^ = (. + „^J^ + (. + 3,^i^t^ = (i + 2) + (I + 3) = 2i + 5. = (x2 + 3x) + (x - 2)(2x + 3) = 3x2 _|_ 2x - 6. 3. If x^ + xy^ + y = 10, dx ^y ^dx 2x + 3xy^f^ + y^ + f^ = 0, whence dy ^ _ 2x + y» ^ dx 3xy» + l' 58 CALCULUS [§41 Exercises Differentiate the following: 1. (x + l)ix - 1). 9. 2/ = (x -h l)VJi^^. 2. (x^ + 2x)(x - 3). 10. y = (2 - x)\/^c^~:^. 3. a;(x2 + 2x - 6). 11. y = {2x ■{■ 3)\/4 - x\ 4. (x - l)='(x^ + 1). 12. ?/ = x\/n^. 5. (x' + 2x - 3)(x + 1)2. 13. xV + 3x - 7?/ = 15 6. xVx - 1. 14. a;2?/ - Sxy^ = lo. 7. (x - 1)V^. 15. yy/^ + xv^ = 3. 8. x(x - l)i 16. xy - xhj = 0. 41. The Derivative of the Quotient of Two Functions. Let u y = -y where u and v are functions of x. Then yv = u. Differentiating by the rule of §40, dv dy _ du ^ dx dx ~ dz du dv dy _ dx ^dx dx V Replacing y by its value, -> (v) du dv Vx ~ ^dx (1) dx The derivative of the quotient of two functions is equal to the denominator times the derivative of the numerator minus the numera- tor times the derivative of the denominator, all divided by the square of the denominator. §42] ALGEBRAIC FUNCTIONS 59 Illustration. 47^72-] _ (^ - 2) -^^ (x^ + 1) -^^ - dx (a; -2)2 _ ix-2)i2x) - (x^+1) (a: -2)2 x^ — Ax — 1 {x - 2)2 Exercises Differentiate the following: 1. ^- 4. ^^-r-'. 7. a;-l 1+a- x - Vx^ - 1 V'x - 1 1 -\-x x^ - 1 Vx x x2 -3 a:2 - 1 g2 4,4 X — 2 V X •'' — 2 „ X + x'^ „ a; „a;3 + 8 3. ~~ 6. T-- 9. hi' 1 - a; (1 - x)^ ^ ~ ^ 42. The Derivative of u", n Negative. In Chapter III the formula dw" , du dx dx was proved for n positive and commensurable. The formula was assumed for negative exponents. We are now in a position to give a proof of the formula for this case. Let y = u-% where s is a positive commensurable constant. Then 1 or yw = 1. Differentiate by the formula of §40, du dy ysu'-^^^+u-^^ = 0, dy sy du = ■ J- = — su- dx u dx 8-1 ^'^ dx (1) 60 CALCULUS [§43 This completes the proof that du° , du dx dx if n is a commensurable constant.^ We see from equation (1) that '(J) nc du dx u°+i dx Illustration. d S 6 dix^ - 1) 12a; (2) dx (x2 - 1)2 (a;2 - 1)» dx {x^ - lY 2. * Exercises llo 4. wmg: 3 X - 1 7 6 . '• x^ + 1 6. 5 (X + ly (x^ + D* 6 3 0. ' . Vx - 1 (1-x^y 1 -X Vx 43. Maximum and Minimum Values of a Fvmction. In Chap- ter I it was shown that the derivative of a function with respect to its argument is equal to the slope of the tangent drawn to the curve representing the function. The derivative is positive where the function is increasing and negative where the function ia decreasing. These facts enable us to determine the maximum and minimum values of a function. Additional exercises in finding maximum and minimum values of a function will be given in this section. Illustration. Let J/ = 2x3 + 33-2 _ 12a; - 10. • ^ = 6x2 + Gx - 12 = 0(3, _j_ 2) (i _ 1). > It can be shown that the formula also holds for incommensurable exponents. 1843] If xi ALGEBRAIC FUNCTIONS 61 If X is less than —2, both factors of the derivative are negative. Then for all values of x less than —2, the derivative is positive and the function is increasing. If x is greater than —2 and less than 1, the first factor of the derivative is positive and the second negative. Hence, if — 2 < x < 1, the derivative is negative and the function is decreasing. If x is greater than 1 the derivative is positive and the function is again increasing. The function changes from an increasing to a decreasing func- tion when x passes through the value —2, and changes from a decreasing to an increasing function when x passes through the value 1. Hence the function has a maximum value when x equals —2, and a minimum value when x equals 1. These values, 10 and — 17, respectively, are obtained by substituting — 2 and 1 for X in the function. (See Fig. 28.) The more important results of the above discussion are put in tabular form below. X X + 2 X - 1 dy dx Function X < -2 + Increasing. -2 < I < 1 + - - Decreasing. 1 < X + + + Increasing. X = -2 — Max. value = 10. X = 1 + Min. value = —17. 62 CALCULUS [§43 It is to be observed that — 2 and + 1 are the only values of x at which the derivative can change sign and that these are the values that need to be examined in finding the maximum and minimum values of the function. Exercises Find where each of the following functions is increasing; decreas- ing. Find the maximum and minimum values if there are any. Sketch the curve representing each function. 1. y ^ xK 6. 2/ = (x + 2)(x -3). 2. y = x\ 1. y = 2a;' - 9x^ + 12x - 10 3. t/ = -2x\ 8. y = x' - 3x + 7. 4. y = x' + 3x - 2. 9. 2/ = x' + x2 - X - 1, 5. 2/ = 3x' - 2x« — 6. 10. y = —^ 11. A sheet of tin 24 inches square has equal squares cut from each corner. The rectangular projections are then turned up to form a tray with square base and rectangular vertical sides. Find the side of the square that must be cut out from each comer in order that the tray may have the greatest possible volume. Hint. Show that the function representing the volume of this tray is 4x(12 — x)'', where x is the side of the square cut out. 12. In a triangle whose sides are 10, 6, and 8 feet is inscribed a rec- tangle the base of which lies in the longest side of the triangle. Ex- press the area of the rectangle as a function of its altitude. Find the dimensions of the rectangle of maximum area. 13. A ship A is 50 miles directly north of another ship £ at a certain instant. Ship B sails due east at the rate of 5 miles per hour, and ship A sails due south at the rate of 10 miles per hour. Show that the distance between the ships is expressed by the function V 125/* — lOOOi + 2500, where t denotes the number of hours since the ships were in the position stated in the first sentence. At what time are the ships nearest together? At what rate are they sepa- rating or approaching when f = 3? When < = 5? When < = 6? 14. The stiffness of a rectangular beam varies as its breadth and as the cube of its depth. Find the dimensions of the stiffest beam which can be cut from a circular log 12 inches in diameter. §431 ALGEBRAIC FUNCTIONS 63 Miscellaneous Exercises Differentiate tlie following twenty-five functions with respect to x: l.i. 9. (3 - x)^ 18. X2(l - X). X 10. (2 - x^y. 19. x(i - x^y. 11. (3 - x=')-2. 20. (1 - X3)2(X -2)2. 4. (X - sy. 12. (2 - x*)-". 13. (x + l)(x - 2). 14. Vi. 21. 22. X - 1. X2 + 1 1 -X 1 +X2 5. (x - 2)^ 15. x\ 23. (x - l)i 6. (x2 - 1)2. x-6) 16. ]. 24. X 7. (x' - 2x2 + Va^ - x2 8. (x - 2) -3. 17. x(x - 1)2. 26. XVI - X2. Integrate the following twenty expressions : 26. ^ = x3. dx 37. '^ .-? ax X i' 27. J = x3+x2. . 38. J = ^^^:^ 28. -^-^ = 0.3 + x2 + X + 1. 39. ^^ ^ ^^ - .c -r u. -r -, -r X. ^^ ^^ _ ^^^^ ^^ dy 1 .^ dy \ — X 29. -7^ = x2. 40. - - dx ~ dx (2x - x2)2 30. ^=-4- 41.^^- ^ dx ^x ' ^^ Vx + 1 31. :7: == a;^ + x^. 42. ■;t- = dx ' ' dx "v/x^-T a.. I = (. - 1,^. 43. g - ^^. 44. —- = x"i dx 46. ^ = X - 1. dx 46. Find ~\i x^ - y"^ = a\ ^-Ftodgitfa + l'-i. 48. A ladder 20 feet long leans against the vertical wall of a 33. dy dx ~ x(l - X2)2. 34. dy dx ~ X2(x3 - 2)5 36. dy dx ~ 1 (1- xy 36. dy dx ~ 1 X2' 64 CALCULUS :§43 building. If the lower end of the ladder is drawn out along the horizontal ground at the rate of 2 feet per second, at what rate is its upper end moving down when the lower end is 10 feet from the wall? Hint. Let AC, Fig. 29, be the wall and let CB be the ladder. I-et AB =x and AC = y. Then and V dy dt = \/400 - But, since dx dt dy dt - X dx \/400 - x^ dt' ■-2X ^400 - x^ The negative sign of the deriva- tive indicates that the upper end pjQ 29. of the ladder is moving down. 49. Answer the question of Exercise 48, ifx=0;x=2;a; = 15; a; = 20. 50. With the statement of Exercise 48, find the rate at which the area of the triangle ABC, Fig. 29, is increasing when the lower end of the ladder is 5 feet from the wall. 51. With the statement of Exercise 48, find the position of the ladder when the area of the triangle ABC, Fig. 29, is a maximum. Fig. 30. 52. A ball is dropped from a balloon at a height of 1000 feet. Ex- press the velocity of its shadow along the horizontal ground as a function of the time, if the altitude of the sun is 20°. Hint. Let x. Fig. 30, be the distance of the falling body above the earth. Let y be the distance of the shadow from a point on the earth directly under the falling body. §43] ALGEBRAIC FUNCTIONS 65 53. With the statement of Exercise 52, find the velocity of the shadow when the ball leaves the balloon; when it is half way to the earth; when it reaches the earth. 64. A man standing on a dock is drawing in a rope attached to a boat at the rate of 12 feet per minute. If the point of attachment of the rope is 15 feet below the man's hands, how fast is the boat moving when 13 feet from the dock? 55. The paths of two ships A and B, sailing due north and east, respectively, cross at the point C. A is sailing at the rate of 8 miles per hour, and B at the rate of 12 miles per hour. If A passed through C 2 hours before B, at what rate are the two ships approaching or separating 1 hour after B passed through C? 3 hours after B passed through C? When are the two ships nearest together? 56. Two bodies are moving, one on the axis of x, the other on the axis of y, and their distances from the origin are given by X = 31^ - 31 + 1, y = 6t - 12, the units of distance and time being feet and minutes, respectively- At what rate are the bodies approaching or separating when 1 = 2? When t = 5? When are they nearest together? 57. A ship is anchored in 35 feet of water and the anchor cable passes over a sheave in the bow 15 feet above the water. The cable is hauled in at the rate of 30 feet a minute. How fast is the ship moving when there are 80 feet of cable out? 58. A gas in a cylindrical vessel is being compressed by means of a piston in accordance with Boyle's law, pv = C. If the piston is moving at a certain instant so that the volume is decreasing at the rate of 1 cubic foot per second, at what rate is the pressure changing if at this instant the pressure is 5000 pounds per square foot and the volume is 10 cubic feet? 59. Water is flowing from an orifice in the side of a cylindrical tan k whose cross section is 100 square feet. The velocity of the water in the jet is equal to \/2gh, where h is the height of the surface of the water above the orifice. If the cross section of the jet is 0.01 square foot, how long will it take for the water to fall from a height of 100 feet to a height of 81 feet above the orifice? 60. At a certain instant the pressure in a vessel containing air is 3000 pounds per square foot; the volume is 10 cubic feet, and it is increasing in accordance with the adiabatic law, piP-* = c, at the rate of 2 cubic feet per second. At what rate is the pressure changing ? 6 66 CALCULUS [§44 61. Water flows from a circular cylindrical vessel whosa radius is 2 feet into one in the shape of an inverted circular cone whose vertical angle is 60°. (a) If the level of the water in the cylinder is falling uniformly at the rate of 0.5 foot a minute, at what rate is the water flowing? (6) At this rate of flow, at what rate will the level of the water in the cone be rising when the depth is 4 inches? When it is 20 inches? 62. A toboggan slide on a hillside has a uniform inclination to the horizon of 30°. A man is standing 300 feet from the top of the slide on a line at right angles to the slide. How fast is the toboggan moving away from the man 3 seconds after leaving the top? 10 seconds after leaving the top? (Use formula for speed of a body sliding down an inclined plane. Neglect friction.) If the man is approaching the top of the slide at the rate of 10 feet a second, answer the same questions, it being supposed that the man is 300 feet away from the top of the slide when the toboggan starts. 44. Derivative of a Function of a Function. If y = <j)(u) and u = fix), ?/ is a function of x. The derivative of y with respect to X can be found without eliminating u. For any set of corre- sponding increments, Ax, Ay, and Au, Ay _ Ay Au Ax Au Ax Hence lim A^ ^ lim 4^ lim Aw^ Since Am approaches zero as Ax approaches zero, dx du dx ^ ' This is the formula for the derivative of a function of a function. Illustration. Let y = u^ -\- 5 and u = 3x^ + 7x + 10. du and §46J ALGEBRAIC FUNCTIONS 67 Then ^ = 3m2(6x + 7) = 3(3a;2 + 7x + 10)^ (6x + 7). dv 45. Inverse Functions. If x = 't>iv),li^ can be found by the rule ^ = A, dx dx dy which is easily proved. dy ^ Mm ^ ^ Mm i_ ^ ±, dx '^^-0 Ax ^x^^Ax dx Ay dy Illustration. U x = 5y^ -\- 7y^ + 3, ^^ = 302/« + 147/, and dy dy 1 dx 2y{lby* + 7) Exercises 1. Find-TT- in terms of X if: ax (a) y = Vu^+ 7 and m = 3a; + 10. (6) 7/ = 2«3 + 5m and i* = x" - 2x. (c) y = — and w = x^ — 2. 2. Find ^ if dx W ^ - (7/2+2)2 3. Find ^ if: dx (a) y* +x* - 7xy = 15. (6) Sxy^ + &x'y + 4x2 = 15. 46. Parametric Equations. If the equation of a curve is given in parametric form, x = f{t), y = (f>{t), it is important to be able to find the derivative of y with respect to x without eliminating « 68 CALCULUS [§47 between the given equations. A rule for doing this can be derived by the method used in §§13 and 17. If I is given an increment At, x and y take on the increments Ax and Ay, respectively. Then Ay Ay At and or Ax Ax At Um At/ lim ^y At=0 ^t M=0 ^x lim Ax At = ^{ dy dy dx dx dt Exercises 1. Find the slope at (6,1) of the curve whose parametric equations arc X = t^ +t, 2/ = < - 1. Find -T- for each of the following: 2. x = t", 2/ = i^ + 1. 3. x = u^ +3, 1 47. Lengths of Tangent, Normal, Subtangent, and Subnormal. In Fig. 31, PT is the tangent and FN is the normal at P. The lengths of the lines PT, PN, TD, and DN are called the tangent, the normal, the subtangent, and the subnormal, respectively, for the point P. Show that the lengths of these lines are; y^ ¥ (1) dx y TD = §47] ALGEBRAIC FUNCTIONS 69 DN = y'£. (2) "^-UHir- (3) dx =4- (I) PN = yjl -h (^] • (4) Exercises 1. Obtain the length of the tangent, normal, subtangent, and sub- normal for the point (1, 2) on the curve y* = 4x. Show that for points on this curve the subnormal is of constant length. • 2. Write the equation of the tangent to y^ = 4x at the pomt (1,2). Write the equation of the normal at the same point. It is to be noted that in this exercise the equations of the tangent and normal lines are to be found, and not the lengths of the tangent and normal as in the preceding exercise. 3. Write the equation of the tangent to I* «* — +— = 1 25 ^9 ^ at the point (3, 2.4). Use implicit differentiation. 4. Find the equation of the curve whose subnormal is of constant length 4 and which passes through the point (1, 3). 2 6. Find the length of the tangent to j/ = - at the point where y = 1. 6. Find the length of the normal to the curve y ■ point where x = 3. 7. Find the equation of the curve passing through the point (1, 3) and having a subtangent equal to the square of the ordinate. CHAPTER V SECOND DERIVATIVE. POINT OF INFLECTION 48. Second Derivative, Concavity. Since the first derivative of a function of x is itself a function of x, we can take the deriva- tive of the first derivative. The derivative of the first derivative is called the second derivotive. In the case of a function y of x, it di^) is denoted by the symbol — -r- — , or -r- vj') , or more commonly by dh/ dhi -j-j* -r-^ is read "the second derivative of y with respect to x." Ifr d^ must here again be remembered that -r-^ is not a fraction with a numerator and a denominator, but is only a symbol representing the derivative of the first derivative. If y = f{x), the first derivative of y with respect to x is some- times written y' and very commonly fix). Similarly the second derivative is indicated by/"(a;). The derivative of the second derivative is called the third deriva- d^v tive. It is designated by -r-^» or ii y = f(x), by f"'{x). The nth derivative is designated by-i-^' or by/^"^^(x). Between the points A and C, Fig. 16, where the curve is con- cave downward, the slope of the tangent decreases from large positive values near A to negative values near C. This means that the tangent revolves in a clockwise direction as the point of tangency moves along the curve from A toward C. Clearly this will always happen for any portion of a curve that is concave downward. (See Fig. 32, a, h, and c.) The slope decreases as the point of tangency moves to the right. On the other hand, if a portion of a curve is concave upward, the slope of the tangent increases as the point of tangency moves to the right. Thus in Fig. 16 the slope of the tangent is negative at C and increases steadily to positive values at E. The same 70 §49] SECOND DERIVATIVE 71 thing is evidently true for any portion of a curve that is concave upward. In this case the tangent line revolves in a counter- clockwise direction. Since the first derivative of a function is equal to the slope of the tangent to the curve representing the function, what has just been said can be stated concisely as follows: If an arc of curve is concave upward the first derivative is an increasing function, while if the curve is concave downward, the first derivative is a decreasing function. If the second derivative of a function is positive between cer- tain values of the independent variable x, the first derivative is an increasing function, the tangent line revolves in a counter- clock- FiG. 32. "wise direction, and consequently the curve representing the function is concave upward between the values of z in question. If the second derivative is negative, the first derivative is a de- creasing function and the curve is concave downward. Thus in Fig. 16 the second derivative is negative between A and C, and between E and G. It is positive between C and E, and between G and /. 49. Points of Inflection. Points at which a curve ceases to be concave downward and becomes concave upward, or vice versa, are called 'points of inflection. At such points the second derivative changes sign. C, E, and G, Fig. 16, are points of inflection. At C, for instance, the second derivative changes from negative values to positive values. Illustration 1. Study the curve y = |x' by means of its derivatives. Differentiating, dx~ ^^* d^y _ 72 CALCULUS [§49 When a; < 0, -^ < 0, -7- = ^x'^ is a decreasing function, and the curve y = |x' is concave downward. When x > 0, d^y dy . . -Y~2 ^ ^> TT ^^ ^^ increasing function, and the curve y = Ix^ is concave upward. Fig. 33. d'^y At the point where x = 0, -7^2 changes sign from negative to positive, and the curve changes from being concave downward to SECOND DERIVATIVE 73 being concave upward inflection. dy Hence the point (0, 0) is a point of Since ^- is positive except when x = 0, ?/ iX* IS an in- creasing function excepting when x = 0. When x = the curve has a horizontal tangent. Fio. 34. In Fig. 33 the graphs of the functions y= \x^ and of its first and second derivatives are drawn on the same axes. Trace out in this figure all the properties mentioned in the discussion. Illustration 2. Let y = Ix' — x^ -\- Ix + 2. Differentiating, J dy _ dx Ix- -2x + l d'y .dx^ h(x X — - l){x - 2. 3). 74 CALCXJLUS [§49 At X = 2, -T^ changes sign from minus to plus. Hence the curve is concave downward to the left, and concave upward to the right of the line x = 2. The point on the curve whose abscissa is 2 is then a point of inflection. The value of the function corresponding to X = 1 is a maximum value, and the value of the function corresponding to x = 3 is a minimum value. See Fig. 34 for a sketch of the function and its first and second derivatives. Trace out in the figure what has been given in the discussion. The more important properties of the function are put in tabular form below. X dx^ dy dx Curve X < 2 X > 2 X = 2 + Decreasing Increasing Concave downward. Concave upward. Point of inflection {y = 2^), X < 1 1 < X < 3 X > 3 X = 1 X = 3 + + Increasing, Decreasing. Increasing, Maximum point (v = 2f). Minimum point {y = 2), Exercises Find the maximum and minimum points and points of inflection of the following curves. Sketch the curves, 1. y = x' — 3x^ 2. y = x' 4- 3x2, 3. 2/ = 2x3 + 3x2 4. ga; _|_ 1, 4. 2/ = 3x* - 4x3 - 1. 6. y = x'. 6. t/ = 2x< - 4x3 - 9x2 _^ 27x + 2, 1. y = 6x^ - 4x3 _|_ I CHAPTER VI APPLICATIONS 60. Area under a Curve: Rectangular Coordinates. An im- portant application of the anti-derivative is that of finding the area under a plane curve. Let APQB, Fig. 35, be a continuous curve between the ordinates X = a and x = b. Further, between these limits, let the curve lie entirely above the X-axis. Our problem is to find the area, A, bounded by the curve, the X-axis, and the ordinates x = a and X = h. The area can be thought of as generated by a moving ordinate starting from x = a and moving to the right to a position DP where the abscissa is x. This ordinate sweeps out the variable area u, which becomes the desired area A when x = b. On moving from the position DP to the position EQ where the abscissa is a; + Ax, the ordinate to the curve takes on an increment Ay and the area u an increment Aw. By taking Ax small enough the curve is either ascending or descending at all points between P and Q. It follows at once from the figure that yAx < Au < (y -{-Ay)Ax, (1) or Au , . y <x:;.<y + ^y- Ax 75 76 CALCUI.US [§50 (If the curve descends between P and Q the signs of inequahty in (1) are reversed. The argument which follows will not be affected.) As Ax approaches zero, Ay approaches zero and y -\- Ay approaches y. Hence v-' which lies between y and y + Ay, approaches y. Thus lim Am ^ Aa:=0 J>^x ^' or If the equation of the curve is y = f(x), g = m. (3) Let F{x) be a function whose derivative is /(x) . Then u = Fix) + C. C is determined by the condition that u = when x = a. Then C = - F(a) and u = F{x) - F(a), (4) an expression for the variable area measured from the ordinate X = a to the variable ordinate whose abscissa is x. A, the area sought, is obtained by putting x = 6 in equation (4). A = F(b) - F(a) (5) Illustration. Find the area A bounded hy y = x', the X-axis, and the ordinates x = 2 and x = 4. du dx = x\ u = \X' ' + c. When X = 2, u = 0, andC = - - A 3* Then and u = \x' _ 8 i) §51j APPLICATIONS 77 Exercises 1. Find the area bounded by the X-axis, the lines x = 1 and x = 2, and (a) y = mx. (b) y = x^. (c) y = 2x2 + 3x + 1. 2. Find the area between the curves y = x^ and x = j/'; . y* = a{a — x) and y = a — x; y^ = ^x and y = 2x; y^ = x* and ?/ = x. 3. Find the area bounded hy y = ■%/ x + 1, the Z-axis, and the ordinates x = and x = 2. 51. Work Done by a Variable Force. In this section there is given a method of finding the work done by a variable force whose line of action remains unchanged. Illustrations of such variable forces are: 1. The force of attraction between two masses, m and ilf, is given by the Newtonian law ItMrn fis) = -^' where s is the distance between the masses and A; is a factor of proportionality. Note that the equation is of the form f(s) - "-,■ 2. The force exerted by the enclosed steam on the piston of a steam engine is, after cut-off, a function of the distance of the piston from one end of the cylinder. 3. The force necessary to stretch a bar is a function of the elongation of the bar. Let AB, Fig. 36, represent a bar of length I, held fast at the left end, A. A force / is applied at its right end and the bar is stretched. It is shown experimentally that up to a certain limit the elongation, s, is proportional to the force applied (Hooke's Law), i.e., f = ks, where A; is a constant depending upon the length of the bar, its cross section, and the material of the bar. 78 CALCULUS [§51 The work done by a constant force in producing a certain displacement of its point of application in its line of action is defined as the product of the force by the displacement. In the problem which we are considering the force varies with the displacement. The work cannot be found by multiplying the displacement by the force. Instead it will be found by integrat- ing an expression for the derivative of the work with respect to the displacement. Let w denote the work done in producing the displacement from s = a to a variable position s = s. Let Aw denote the work done in producing the additional displacement As. Let / denote the force acting at s, and / + A/ the force acting at s + As. A/ I I FiG. 36. may be positive or negative according as the force increases or decreases with distance. For definiteness suppose A/ positive. In producing the displacement As the force varies from / to / + A/, and hence the work Aw; lies between /As and (/ + A/) As, which represent the work which would have been done had the forces / and / + A/, respectively, acted through the distance As. Hence /As < Aw; < (/ + A/) As, or f<fs<f+^f- As As approaches zero, A/ approaches zero, and we obtain, dw ds = f. (1) §51] APPLICATIONS 79 Integration gives w = F{s) + C, (2) where F{s) is a function whose derivative is /. When s = a, w = Q, (2) gives C = — F{a) and, w =F{s) -Fia). (3) This represents the work done in the displacement from s = a to s = s. The work, W, done in the displacement from s = a to s = & is obtained by substituting 6 for s in (3). W = F(b) - F(a). (4) Illustration 1. Find the work done in stretching a spring from a length of 20 inches to a length of 22 inches, if the length of the spring is 18 inches when no force is applied and if a force of 30 pounds is necessary to stretch it from a length of 18 inches to a length of 19 inches. Denote the elongation of the spring by s. In accordance with Hooke's Law, / = ks. Since s = 1 when / = 30, A; = 30 and/ = 308 Substituting in equation (1), -r = 30s. as w = 15s2 + C. The problem is to find the work done in changing the elonga- tion from s = 2 to s = 4. When s = 2,w = 0. Hence C = — 60, and w = 15s2 - 60. The required work, W, is found by giving to s the value 4. TF = 240 - 60 = 180. Thus the work done is 180 inch-pounds, or 15 foot p junds. Illustration 2. Two masses M and m are supposed concen- trated at the points A and B, respectively. Find the work done 80 CALCULUS [§51 against the force of attraction in moving the mass m along the line AB from a distance a to a distance b from the mass M, the latter mass being fixed. If / is the force of attraction between the two masses, _ kmM f - g2 Then dw _ kmM Is ~~ s2 From which kmM w = — + C. -S +AS- FiG. 37. When s = a, w = 0. Hence kmM and w = kmM — \_a 1- To find the required work, PF, let s = 6, W = kmM - - [a 1- b_ Illustration 3. Gas is enclosed in a cylinder, one end of which is closed by a movable piston. Find the work done by the gas in expanding in accordance with the law pv^-* = K, from a volume of 3 cubic feet at a pressure of 15,000 pounds per square foot to a volume of 4 cubic feet. Let A be the area of the cross section of the cylinder. Then pA is the force on the piston, Fig. 37, and dw ds=P^' §51] APPLICATIONS or dw AK AK K I ds ~ v^* ~ (AsY* ~ A'^U^*' Integration i gives 1 K 1 ^~ 0.4 ^""s"'*'^^ - ^ ^ +C When V = 3 ,w = 0. Hence 1 K 0.4 3°"' and K r 1 1 -] ^ 0.4 [s^* v°*y 0.4 [3°" 40''_ 81 When t; = 3, p = 15,000. Hence K = (15,000)(3i"), and W = 150,000[0.75 - (0.75) i«l = 12,230. Exercises 1. A spring is 12 inches long and a force of 120 pounds is necessary to stretch it from its original length, 12 inches, to a length of 14 inches. Find the work done in stretching the spring from a length of 13 inches to a length of 15 inches. 2. In the case of a bar under tension, Fig. 36, the relation between the stretching force, /, the original length of the bar, I, and the elonga- tion of the bar, s, is given by ^ EAs ^ = -r' where E is the modulus of elasticity of the material of the bar and A is the area of the cross section of the bar. Find the work done in stretching a round iron rod ^ inch in diameter and 4^ feet long to a length of 54.5 inches, given that E = 3-10^ pounds per square inch. 3. A spherical conductor, A, is charged with positive electricity and a second spherical conductor, B, with negative electricity. The force of attraction between them varies inversely as the square of the 6 82 CALCULUS [§52 distance between their centers. If the force is 10 dynes when the centers are 100 centimeters apart, find the work done by the force of attraction in changing the distance between the centers from 140 centimeters to 120 centimeters. 4. Find the work done by a gas in expanding in accordance with the law pv^ = C from a volume of 5 cubic feet to one of 6 cubic feet, if p = 70 pounds per square inch when v = 5 cubic feet. 5. Find the work done in compressing a spring 6 inches long to a length of 5§ inches if a force of 2000 pounds is necessary to compress it to a length of 5 inches. 6. The work done by a variable force can be represented graphically as the area under a curve whose ordinates represent the force. Con- struct the figures and prove this fact for Illustrations 1, 2, and 3. 62. Parabolic Cable. Suppose a cable, AOB, Fig. 38, a, is loaded uniformly and continuously along the horizontal, i.e., so that any segment of the cable sustains a weight proportional to the projection of the segment upon a horizontal line. Let k be the weight carried by a portion of the cable whose horizontal pro- jection is one unit of length. Choose 0, the lowest point of the cable, as origin and a hori- zontal line through as axis of x. Let P be any point on the cable. Suppose the portion OP of the cable cut free, Fig. 38, b. To keep this portion in equilibrium a horizontal force H and an inclined force T must be introduced at the points and P, respectively. The force H must be equal in magnitude to the tension in the cable at 0, and it must act in the direction of the tangent line at that point. Similarly, the force T must be equal to the tension in the cable at the point P and act in the direction of the tangent line. The force T can be resolved into its vertical and horizontal §53] APPLICATIONS 83 components V and H', respectively. Now H and H' are the only horizontal components of the forces acting on OP and, since OP is in equilibrium, they must balance each other. Therefore, H = H'. (1) Hence the horizontal component of the tension in the cable is independent of the point P, i.e., it is a constant. In like manner the only vertical components of the forces acting on OP are the weight kx supported by OP, acting downward, and V, the vertical component of T. They must balance one another. Hence V = kx. (2) V The slope of the tangent Ime to the curve at the point P is „/ Then dy _ V^ _kx dx~ H'~ H' ^'^' This is the slope of the curve at any point. On integrating we obtain the equation of the curve apart from the arbitrary constant C. y = % + c. (4) C is determined by the condition that y = when x = 0. Then C = 0, and (4) becomes y = 2^- (5) This is the equation of a parabola with its vertex at the origin. 63. Acceleration.^ In §38 acceleration was defined as the time rate of change of velocity, i.e., as the derivative of the velocity with respect to the time. But velocity is the derivative of dis- tance with respect to time. Hence the acceleration is the second, derivative of the distance with respect to the time. If s denotes the distance and t the time, the acceleration is expressed by -w-. In the case of a freely falling body d'^s „. • The statements in this section refer to motion in a straight line. 84 CAIiCULUS [§53 The relation between s and t can be found from this differential equation by integrating twice, as follows: The first integration gives and the second s = igt^ + Cit + C2. (3) Two arbitrary constants of integration are introduced. They can be determined by two conditions. If s = So (4) and v = -^ = Vo (5) when t = 0, (2) gives Ci = Vq, and (3) gives C2 = Sq. Then s = hgt' + vot + So. (6) This result was found in §38 by essentially the same method, where dv d^s the symbol n^ was used instead of -n^' Exercises 1. Solve Exercise 5, §38, by the method used above. 2. Obtain the relation v = \/2gs (see §38) directly from the equation ds Hint. Multiply by 2 -jr • dt^ = 9- ds d^s _ n f^s ^dt di^ ~^^dt' (ds\i ~T7 I with respect to t and the second that of 2gs. We then have Determine C by the condition that v = Q when 8 = 0. §54] APPLICATIONS 85 64. The Path of a Projectile. An interesting application of integration is to find the equation of the path of a projectile, a baseball for instance, thrown with a given velocity at a given inclination to the horizontal. Let 0, Fig. 39, be the point from which the ball is thrown. Take this point as the origin of a system of rectangular coordi- nates. Let the ball be thrown so that its direction at the instant of leaving the hand makes an angle a with the horizontal, and let the initial velocity of the ball be Vq. Then the horizontal com- ponent of the initial velocity is Vo cos a, while its vertical component is Vo sin a. That is, at the instant the ball is thrown its x-coordinate is increasing at the rate of Vo cos a feet per second. Similarly the initial rate of change of the ^/-coordinate is vq sin a. Fig. 39. At the end of t seconds after the ball was thrown it is at the point P whose coordinates are x and y. If the resistance of the air is neglected there is no force acting on the ball tending to change the component of its velocity parallel to the X-axis. Hence the x-component of the velocity is at all times the same as at the beginning, viz., Vq cos a. The x-component of the velocity is also dx -rr, viz., the time rate of change of the abscissa of the ball. There- fore we can write dx . . jT = Vo cos a. {1) From which on integration X = {vo cos a)t + C. (2) Time is counted from the instant the ball was thrown. The 86 CALCULUS [§54 condition for determining C is then that a; = when f = 0. It follows that C = 0, and (2) becomes X = {vo cos a)t. (3) This equation gives the x-coordinate of the ball at any time t. In the vertical direction, the force of gravity acts to change the t/-component of the velocity. Then g= -.. (4) The negative sign is used since the force of gravity causes the velocity in the direction of the positive F-axis to decrease. In- tegration gives 1= -gl + C. (6) dv d is determined by the condition that 37 = ^0 sin a when t = 0. Then C2 = Vq sin a and (5) becomes ^y = — gt + v^ sin a. (6) Integrating again, y = - W + i^o sin ci)t + C3. (7) Since y — when ^ = 0, C3 = 0, and (7) becomes y = — hgi^ + (^0 sin a)L (8) This is the ^/-coordinate of the ball at any time t. Equations (3) and (8) are the parametric equations of the path of the ball. The elimination of t between these equations gives the rectangular equation of the path, y = — :r~2 r~ x^ + x tan a. (9) " 2^0 cos^ a This is the equation of a parabola with its vertex at the point, r?;o^ sin 2a Vq'^ sin^ a"! L 2g ' ~2^J" Tt is to be remembered that in the solution of this problem the §54] APPLICATIONS 87 resistance of the air was neglected. Consequently the results obtained can be regarded only as approximations. Experimentally it has been shown that the resistance of the air increases with the velocity of the moving body. For low velocities the resistance is assumed to vary as the first power of the velocity, but for higher velocities, such as are attained by rifle balls, the resistance is as- sumed to vary with the second power, and the results obtained above cannot be considered to be even approximations. Exercise 1. Find the angle of elevation, a, at which the ball must be thrown to make the range, OA, Fig. 39, a maximum. CHAPTER VII INFINITESIMALS, DIFFERENTIALS, DEFINITE INTEGRALS 55. Infinitesimals. In §23 an infinitesimal was defined as a variable which approaches the limit zero. Thus, x^ — 1, as z approaches 1, is an infinitesimal. It is to be noted that a variable is thought of as an infinitesimal only when it is in the state of approaching zero. Thus x^ — 1 is an infinitesimal only when x approaches +1 or —1. An in- finitesimal has two characteristic properties: (1) It is a variable. (2) It approaches the limit zero; i.e., the conditions of the problem are such that the numerical value of the variable can be made less than any preassigned positive number, however small. This meaning of the word infinitesimal in mathematics is entirely different from its meaning in everyday speech. When we say in ordinary language that a quantity is infinitesimal, we mean that it is very small. But it is a constant magnitude and not one whose numerical measure can be made less than any preassigned positive number, however small. Thus, 0.000001 of a miUigram of salt might be spoken of as an infinitesimal quantity of salt, but the number 0.000001 is clearly not an infinitesimal in the sense of the mathematical definition. On the other hand, if we have a solution containing a certain amount of salt per cubic centimeter and allow pure water to flow into the vessel containing the solu- tion while the solution flows off through an overflow pipe, the quantity of salt per cubic centimeter constantly diminishes. The amount of salt left in solution after a time t is then an infinitesimal, as t becomes infinite. Infinitesimals are of fundamental importance in the Calculus. The derivative, which we have already used in studying functions, is the limit of the ratio of two infinitesimals. Ay and Ax. 56. ^f^ • Let the arc AB, Fig. 40, subtend an angle a 88 INFINITESIMALS 89 at the center, O, of a circle of radius r. The angle a is measured in radians. Let AT he tangent to the circle at A, and let BC be perpendicular to OA. The area of the triangle OCB is less than the area of the circular sector OAB, and this in turn is less than the area of the triangle OA T. UBC){OC) < har^ < ^iAT)r BC OC ^ ^AT — — - < a < r r r OC . ■ — sin a < a < tan a r OC < sin a As the angle a approaches zero, OC OC approaches r and — approaches 1, and further, cos a approaches 1. Hence the first and last members of the inequalities (2) approach the same limit, 1. Then the second a member, -; — ■» which lies between ' sin a them, must approach the same limit, 1. Therefore lim sin a lim a=0 1 a sin a = I. (3) 57. lim tan a lim tana sin a tan a a=0 lim a ^ lim /sina 1 \ «-o \ a cos a/ ^ Aim sina\ /lim 1 \ \a=0 cc / \°'=° cos OC/ = 1. lim tan a lim a=0 = 1. (1) (2) 58. lira 1 — cos a 1=0 90 CALCULUS [§59 Since ^ 2 sin'' fT sin 7, 1 — cos a 2 2 . a = = sin ^» a a a 2 lim l-coso! /lim ^^" 2 W IJm . a cc^Q a ~ I «=^o a / I «^o ^ 2 Hence lim 1 — cos « _ a=^o « ~ "• ^^^ In Fig, 40, AB, AC, AT, BT, and BC are infinitesimals as a approaches zero. Then, from (3), §56, i™ ^ = h from (1), §57, i^o ^ = 1, from (2), §57, i^o ^^ = 1, from (1), §58, i^o J^ = 0. 59. Order of Infinitesimals. Consider the infinitesimals x^ and X as X approaches zero. The ratio of x^ to x is x, which is itself an infinitesimal. The infinitesimals x^ and x are repre- sented. Fig. 41, by the ordinates MP and MN, to the curves y = x^ and y = x. The quotient X ~ MN is a measure of the relative magnitude of these infinitesimals as they approach zero. It shows that MP becomes small so much more rapidly than MN that the limit of their quotient is zero. On the other hand, the infinitesimals 2x and x behave very 2x differently. Their quotient is — = 2, and the limit of this J59] INFINITESIMALS 91 quotient is 2. In this case the limit of the ratio of the infini- tesimals is not zero. (See Fig. 42.) Again, lim 1 — c os a while = 0, lim sma «-0 rv = 1. These illustrations of the comparison of two infinitesimals lead to the following definitions of the order of one infinitesimal with respect to another. Two infinitesimals, a and /3, are said to be of the same order if the ex limit of ^ is a finite number not zero. Fig. 41. Fig. 42. 7/ the limit of ^ is zero^ a is said to be of higher order than jS. Thus, 2x and x are of the same order; x^ is an infinitesimal of higher order than x; sin a and a, or CB and AB, Fig. 40, are of the same order; tan a and a, or AT a,nd AB, Fig. 40, are of the same order; tan a and sin a, or CB and AT, Fig. 40, are of the same order; 1 — cos a is of higher order than a, or CA, Fig. 40, is of higher order than AB. Let ACB, Fig. 43, be a right angle inscribed in a semicircle. Let BD be a tangent line, and let CE be perpendicular to BD. If 92 CALCULUS [§59 the angle CAB approaches zero, BC, CD, BE, ED, CE, and arc BC are infinitesimals. From similar triangles AB BC BD AC~ BE~ BC' AB Since lim a^— 1, it follows that ,. BC .. BD ^ Hence BC and BE, and BD and BC are infinitesimals of the same order. Again, BC CE CD AB BC ~ BD' D E ,• BG ^ Smct! lim -ir-f, = 0, AB ' Fig. 43. Hence CE is an infinitesimal of higher order than BC, and CD is an infinitesimal of higher order than BD. Again, CB ^CE^ _CD AC ~ BE ~ CB Since lim -j-^ = 0, AC ' hm^ = lim^ = 0. Hence CE is an infinitesimal of higher order than BE, and CD is an infinitesimal of higher order than CB. §60] INFINITESIMALS 93 Exercises 1. Show that X — 2x^ and 3a; + x^ are infinitesimals of the same order as x approaches zero. 2. Show that 1 — sin d and cos'' 9 are infinitesimals of the same order as 6 approaches g- 3. Show that 1 — sin is an infinitesimal of higher order than cos as 6 approaches ^^ 4. Show that sec a — tan a is an infinitesimal as a approaches -• 5. Show that 1 — sin a is an infinitesimal of higher order than sec a — tan a as a approaches ^• 6. Show that 1 — cos 6 is an infinitesimal of the same order as 0^ as 6 approaches zero. T ou +1, + lim sin g - g 7. Show that ^^q = 0. o ou +1. + lim sine - 9 8. Show that „ . „ — . — r— = 0. »=o sin 9 9. Show that i™ *^^^T^^ = °- 10. Show that 1™*^J^^ = 0. *=o tan d ^^ n, . > , lim sin a — tan a 11. Show that .^ 7 = 0. «=0 tan a 12. Show that^'"^ /^".-^^"" = 0. «=o sin a 60. Theorem. The limit of the quotient of two infinitesimals, a and /3, is not altered if they are replaced by two other infinitesimals, a B y and 5, respectively, such that lim -- = 1 and lim -: = 1. Proof : a y ~ a _ 7 ~B~ TT' lim — lim ^ = a lim -i = lim j i hm g^ 94 CALCULUS [§60 since lim — = lim -v = 1- 7 It is evident from the proof that the limit of the quotient is unaltered if only one of the infinitesimals, say a, is replaced by T another infinitesimal 7, such that lim - = 1. a Illustrations. L Since lim sin a _ a=0 ~ ~ ^» a lim 1 ~" cos a _ Hm 1 ~ cos a «-o sin a ~ "=o q. = 0. 2. Since lim tan a a = a 1, lim 1 — cos « _ lim 1 — cos « _ «-0 tan a "-0 a 3. In Fig. 40, lim ^ ^ lim C;4 _ Urn C'A ^ cc^oAB »=OBC "=0AT Exercises l.Showthat^^"l ^"7" = i Hint. 1 — cos a = 2 sin^ ^- 2. Show that lim ^i^«a--cos«) ^ a=0 a^ ^ 3. Show that lim («-5)-sina ^ ^5. a=0 a 4. Show that"'" ^"'"'« ' «=Ocos a sin- a ^* K «, ,, . lim 3 3:" - 4x3 5. Show that ;,^0 2 x^-5x< = ^' Hint. Replace numerator by 3x* and denominator by 2x*. A .A A 6. Show that "."^^—4^= ^^-^ = 2. X^ X^ X^ 1611 DIFFERENTIALS 95 61. Diflferentials. Let PT, Fig. 44, be a tangent line drawn to the curve y = f{x) at the point P. Let DE = Ax, RQ = A?/, and let angle RPT = t. From the figure, RM . .,. . = tanr = / (x), Ax or RM = fix) 'Ax. This is the increment which the function would take on if it were to change uniformly at a rate equal to that which it had at P. This quantity, j'{x)Ax, is called differential y, and is de- noted by dy. Its defining equa- tion is dy = f'{x)Ax. (1) Ax, the increment of the independent variable, is called differential X and is denoted by dx, i.e., Ax = dx. Equation (1) becomes dy = j'{x)dx\ (2) In Fig. 44, RM = dy and DE = PR = dx. In general, dy is not equal to Ay, the difference being MQ, Fig. 44. However, it will be shown that ^^^g ^ — ~ ^• lim RQ_f,(^s or lim \RQ_ Rm = .>r^. ix=0 RM PR J ^•^^' ^^^0 IRM PRJ (3) RM . But, since p^ is constant and equal to/'(x), equation (3) becomes lim RQ Ax=0 RM 1, > In the expression (2) for the differential of the function /(i), the first derivative is the coefficient of the differential of the argument, and for this reason it is sometimes called the differential coefficient. 96 CALCULUS (§62 at It is to be noted that dx is an arbitrary increment and that dy is then determined by this increment and the value of the deriva- tive, i.e., by the slope of the tangent at the point for which the differential is computed, dx and dy are then definite quantities and we can perform on them any algebraic operation. Thus we can divide (2) by dx and obtain t = f'M. (5) where dy and dx denote the differentials of y and x, respectively. Thus from the definition of differentials the first derivative may be regarded as the quotient of the differential of y by the differ- ential of X. It is to be observed, however, that this statement gives no new meaning to the derivative, since the derivative was used in the definition of the differential. 62. Formulas for the Differentials of Functions. In accordance with equation (5) of the preceding section, any formula involving first derivatives can be regarded as a formula in which each first derivative is replaced by the quotient of the corresponding differentials. Thus, , /m\ du dv dx dx dx Each derivative being considered as a fraction whose denominator is dx, we can multiply by dx, and obtain vdu — udv e) = v In words, the differential of a fraction is equal to the denominator times the differential of the numerator minus the numerator times the differential of the denominator, all divided by the square of the denomi- nator. It will be noted that the wording is the same as that for the derivative of a fraction except that throughout the word differential replaces the word derivative. §62] DIFFERENTIALS 97 The other formulas for derivatives which have been de- veloped are expressed below with the corresponding formulas for differentials. Formulas 1. dc dc = 0. 2. d{cu) du dx ~ dx d{cu) = cdu. 3. d{u + v) du dv dx ~ dx dx d{u + v) = du + dv. 4. du'* , du dx dx dw* = nM"~^ du. 6. d(uv) dv du dx dx dx d(uv) = udv -{- vdu. 6. , /u\ du dv a ) v~j wt— \v 1 dx dx dx ~ v^ , /u\ vdu — udv 7. \v / dx dx ~ v^ , /c \ cdv 8. dx v*^ / c \ cndv 9. du du dx dJ - ^" . 2u^ The formula for the differential oi y = cu" can be put in the fol- lowing convenient form : (fy du 10. — = n — ■} y u which is obtained directly by dividing dy = cnu'*~^du hy y = cW. The process of finding either the derivative or the differential of a function is called differentiation. The process of finding a function when its derivative or differential is given is called integration. We have no symbol representing integration when applied to derivatives. The symbol for integration when applied to dif- 98 CALCULUS [§62 forentials is j . Thus I Sx^dx = x^ -\- C. The origin of this symbol will be explained later. It is read "integral of." Illustrations. 1. U y = Vn^S dy = Ui - x2)-5( - 2xdx) _ xdx Vl - x^' By formula 10, where w = 1 — x^, dy _ 1 — 2x dx J ~2 (1 - a;*) xdx 2. If 2/ = 1 -a;* X X2- 1 ^y = (^^^^Tp _ (x' — 1) c?a; — X (2a; dx) ~ (x2_ 1)2 ■ ^ _ (x^ + l)dx (x* - ly ' 3. If d?/ = xdx, y = ixdx = A r2xdx =i-+c. 4. If dy = xVl — x^ dx, y = Cx{l — x^)^dx = -H/HI -x2)2( _2xdx) (l-x^)^ + C. §62] DIFFERENTIALS • 99 dtj dx r' 2/ = C(x - 1). 5. If - , y X - 1 by formula 10. ft If d]i xdx y x^ — 1 dy _ 1 2xdx y ^ 2 x2- 1 y = C\/a;2 - 1. Exercises Find dy in the following ten exercises : 1. J/ = x2 - 3x - 2. ^ V: 6. y = 2. 2/ = (x-l)» X- 1 7. J/ = (x - l)(x» - 1)2. 3. y = x^- x~^ - 3a;. 8. y = {x* + x - 2)». ,_i 4. y = (x - 2)*. 9. y = (x - 1)" 6. y = (x2 - 2)i 10. y = (x2 - 1)~^. Integrate the following: 11. fxMx. 12. C{x^ - l)xdx. 13. fCx' - 3x + 5) (x2 - 1) dx. 14. r(x» - 2x - 6)3 (x - 1) dx. /dx 16. Cs/x dx. "■ /t 18. Cx^dx. 100 CALCULUS m 19. dy y = X 20. dy _ xdx x2- 1 21. dy _ (x^ - 2x + 4)dx y ~ a;3-3x2+12a;-2' lim Ax Since^ 63. Differential of Length of Arc : Rectangular Coordinates. Let PR, Fig. 45, = Ax, RQ = Ay, the chord PQ = Ac, and the arc PQ = As. (s represents the length of arc measured from some point A.) PT is the tangent at P. (Ax)2 + {Ayy m /^\2 _ 1 , lim (^Y__ 1 , (f^vy ^0 \Ax) ^ "^ ^^=^0 \^xj ~ ^ "^ \dxl lim Ac A2:=0 ^s = 1, > When Ai is taken so small that the curve has no point of inflection between P and Q, the chord PQ < arc PQ < PT + TQ, or Ac < As < Pr + TQ. Whence, Therefore Then from (1), i<^<^+m (ptX 2 ^ (da:) 2 + (dy)2 ^ [^ \ dx) [ac J (Ax) 2 + (A2/)2 J _|_ /^\f lim /• Ax=0 I Ac I = 1. lim TQ Ax=0 -^ lim Aj/ — dy Ay ^^-0 A^ a;^ r lim /. ''i/^l r 'im ^ l „ n = [Ax^oV^ - Ayjj [ax=0 Ac J "' lim ^ =, Ax— Aj/ lim As Ac Aa;=s=0 (1) DIFFERENTIALS 101 Ac can be replaced by As ( §60) . lira /As\ 2 ^ , , /M ' ^=0 \Ax) "^ [dxj ' or (ds)2 = (dx)2 + (dy)2 (1) ds=Vl+(g)'dx (2) ds = Vi + a~)^<iy Vdy/ - (3) Equation (1) shows that the line PT, Fig. 45. represents ds. If T denotes the angle made by the line PT with the positive X-axis, dx = cos T ds dy = sin r ds. Illustration. Find the length of the curve y = fx^ between the points whose abscissas are 3 and 8. ^= x^ dx (I)'- Substituting in formula (2), ds = Vl + X dx. Integrating, s= 1(1 + x)' +C. When X = 3, s = 0. Hence C = - Y, and s= iil + xy -W This formula gives the length of the curve measured from the point whose abscissa is 3 to the point whose abscissa is x. On placing X = 8 we obtain s = ^3^, the length of the curve from 102 CALCULUS I §64 the point corresponding to x = 3 to the point corresponding to X = 8. Exercises Find the differentials of the length of the following curves: 1. y = x'. 2. x^ + y^ = 4. 3. 2/ = xK 4. 2/2 = X. 5. 3x2 + 4y2 = 12. 6. xy = 1. 7. xj/2 = 1. 8. 7/ = x-^ 64. The Limit of 2f(x)Ax. Let y = /(x) be a continuous function between x = a and x = 6. In §50 it was shown that the area bounded by the curve, the X-axis, and the ordinates X = a and x = 6 is given by the formula A = F{b) - F{a), (1) where F{x) = J f(x)dx. A second expression will now be found for the area. Divide the interval h — a, Fig. 46, into n equal parts and at each point of division erect an ordinate. Complete the rectangles as indicated in the figure. !/=/(«) The sum of the rectangles of which DEQ'P is a type, is approxi- mately equal to the area ABUV. The greater n, the number of rectangles, i.e., the smaller Ax, the closer will the sum of the rectangles approximate the area ABUV. We say then that _ lim 1"^ 2 DEQ'P, §65] DIFFERENTIALS 103 or x = b The above expression (2) represents the actual area and not an approximation to it, as can be shown by finding the greatest possible error corresponding to a given number of rectangles and then proving that this error approaches zero as the number of rectangles becomes infinite. Thus it is easily seen that the difference between the true area A and the sum of the rec- tangles is less than the area of the rectangle RSTU. The altitude, f{b) — f{a), of this rectangle is constant while the length of the base, Ax, approaches zero. Hence the area of RSTU approaches zero. Therefore the limit of the sum of the rectangles is the area sought. On equating the two expressions for A, given by (1) and (2), we have x = b where Fix) = ff(x)dx. This equation is the important result of this section. It gives a means of calculating Ax= X =a For, to calculate this limit we need only to find the integral of f{x)dx and take the difference between the values of this integral at X = a and x = h. The result of this section will be restated and emphasized in the next section. 65. Definite Integral. The expression x = a which was introduced in the preceding section is of such great 104 CALCULUS [§65 importance that it is given a name, "the definite integral of f{x) between the limits a and b," and is denoted by the symbol i f{x)dx. Equation (3), §64, gives a means of calculating the value of the definite integral. The function F(x), the integral oi f{x)dx, is called the indefinite integral of j{x)dx in order to distinguish it from the definite inte- gral which is defined independently of it, viz., as the limit of a certain sum. We have then the following definition and theorem: Definition. Let J{x) be a continuous junction in the interval from X = a to X = b, and let this interval be divided into n equal parts of length Ax by points Xi, Xz, Xi, . . ., x„_i. The "definite integral of f{x) between the limits a and b" is the limit of the sum of the products f{xi) Ax formed for all of the points Xq = a, Xi, xj, . . . , x„_i, as the number of divisions becomes infinite. Theorem. The definite integral of f(x) between the limits a and b is calculated by finding the indefinite integral, F(x), of f{x)dx and forming the difference F{b) — F{a). The symbol for the definite integral, fix)dx, r is read "the integral from a to b of f{x)dx." As we have seen, it means h Jf{x)dx = J2i Zff{x)Ax. a a Many problems, such as finding the work done by a variable force, the volume of a solid, the coordinates of the center of gravity, lead to definite integrals. But, no matter how a definite integral may have been obtained and no matter what other meaning it may have, it can always be regarded as repre- senting the area included by the curve y = f(x), the X-axis, and the ordinates x = a and x = b, provided that /(x) is a function DIFFERENTIALS 105 which can be represented by a continuous curve. This fact, that f{x)dx r can be regarded as representing an area, enables us to calculate its value. For the area in question is equal to F{b) — F{a), where F{x) is the indefinite integral of f{x)dx. Consequently we have, in all cases, f{x)dx = F{b) - F{a). r This is often written X b j{x)dx = F{x) = F{b) - Fia), to show how the result is to be calculated. Thus I x^dx = -^ 2^ _ P _ 7 3 3 ~ 3' Exercises Evaluate the following definite integrals: r C'dx C 1. I (2x + Z)dx. 2. I ^- 3. I Va^ + x" xdx. Ji Ji Jo 66. Duhamel's Theorem. If ai, ocz, as, • ■ •, oin are n in- finitesimals of like sign, the limit of whose sum is finite as n becomes infinite, and if /3i, jSz, /Ss, • • , /3„ are a second set of infinitesimals such that lim §J _ 1 n=co ^. - A. where i = 1, 2, 3, • • • , n, then lim V „.. = lim V p,. n=m ^ n=co ^ t^*' 1 1 Proof. Let — = 1 + €<. OCi 106 CALCULUS [§66 Since lim ^i _ . n=co „. A. ^^ €i = 0. n=oo At first let it be assumed that the a's are positive. Let E be the numerical value of the largest e, i.e., E >. le<', i = 1, 2, 3, • • • , n. Then, since /3i = «» + e.a,, i = 1, 2, 3, • • • , n, ai — Eai < ^i £ ai + Eai az — Ea-i <. ^1 ^ a2 + Ea^ an - EUn <^n<an-\- E Un- Adding, we get i = n x=n »' = » {l-E)%ai<^^i<{l-\-E) X cii. 1=1 i = 1 »■ = 1 ^^E = Q, n=oo n i = n lim ^ a< = lim ^ /3i t = 1 » = 1 and the theorem is proved. If the a!s are negative, it will be necessary to change the proof just given, only by reversing the signs of inequality. Section 64 furnishes an illustration of this theorem. In this example the limit of the sum of the infinitesimal trapezoidal areas DEQP is finite as n becomes infinite, since it is the area sought. DEQ'P < DEQP < DEQP', (see Fig. 46), or yAx < DEQP < {y + Ay) Ax, or DEQP y + Ay _ ^ ^ DEQ'P ^ y This shows that the limit of the ratio of the trapezoidal area to Since n=< t = n §67] DIFFERENTIALS 107 the area of the corresponding rectangle is 1 as n becomes infinite. Then by Duhamel's Theorem, Um 2 DEQ'P = ^^^ 2 DEQP = A, Since we are able to replace the infinitesimals DEQP by the infinitesimals DEQ'P, we may calculate the area which is the sum of these infinitesimals by means of the definite integral. This is a characteristic process in the use of the definite integral. The quantity sought is subdivided into n portions which are infinitesimals as n becomes infinite. These are replaced by n other infinitesimals of the form f{xi) Ax. The limit of the sum of the latter infinitesimals is a definite integral. Since the limits of the two sums are equal by Duhamel's Theorem, the definite integral is equal to the quantity sought. Illustrations of the applications of Duhamel's Theorem to obtain definite integrals representing work, force, volume, etc., follow. 67. Work Done by a Variable Force. In §51 there was found the work done by a variable force, /(s), in producing a displacement Fig. 47. from s = a to s = &. We shall now obtain the same result by building up the definite integral which represents the work. Divide the total displacement h — a, Fig. 47, into n equal parts of length As. The force acting at the left end of one of these parts is /(s), while that acting at the right end is /(s + As). The total work done in producing the displacement, 6 — a, is approximately 8 = 6 s = a The actual work is the limit of this sum as As approaches zero.^ ' This step can be justified by using Duhamel's Theorem. Let Au) represent the work done in producing the displacement As. Then 108 CALCULUS [§67 Illustration 1. The solution of the problem of Illustration 1, §51, is expressed by 4 I SOsds = 15s2 = 180. 2 Illustration 2. The solution of the problem of Illustration 2, §51, is expressed by w = kmM I — - = — kmM — = — kmM \r I J„ s- s „ L6 a J = kmM\ Illustration 3. In solving the problem of Illustration 3, §51, we can write _ lim « = o where t)2 and t^i are the volumes corresponding to s = a and s — b, respectively. Since pv'' = C, p = ^' and , C'dv _ C J C The student will complete the numerical work. n= 00 But /(s)As < Au) < /(s + As)A8, , ^ Aw /(s-fAs) °' ^</(.)A«<-7(«) Then lim Am 1 _ A; ^'"'~'' ~ '"'~''^- Hence by Duhatnel's Theorem b As=0 /(s)A8 So 2^-= ii"o 2/WA.=/j'/(,M.. o a " s = b c s = a *f "■ DIFFERENTIALS 109 Exercises 1. Set up and evaluate definite integrals representing the work sought in Exercises 1-5, §51, Chapter VI. 2. Water is pumped from a round cistern whose median section is a parabola. The cistern has a diameter of 8 feet at the top and it is 16 feet deep. The water is 10 feet deep. Find the work done in pumping the water from the cistern if the discharge of the pump is 3 feet above the top of the cistern and if the friction in the pump and the friction of the water in the pipes are neglected. 3. Find the work done by a gas in expanding in accordance with the law pv^-* = C from a volume of 10 cubic feet to one of 12 cubic feet, if when V = 9 cubic feet p = 100 pounds per square inch. \B 4. Find the work done in stretching a spring whose original length was 15 inches from a length of 16 inches to a length of 18 inches if a force of 40 pounds is required to stretch it to a length of 16 inches. 6. Find the work done in compress- ing a spring of original length 5 inches to a length of 3§ inches, if a force of 900 pounds is required to compress it to a length of 4 inches. 6. The force due to friction is pro- portional to the component of force Fig. 48. normal to the surface over which a body is being moved. Find work done in dragging a body weighing 100 pounds from the base to the top of a slide in the form of a segment of a sphere, Fig. 48, if the distance AB = 200 feet and the radius of the sphere is 500 feet. Express the result in terms of /*, the coeffi- cient of friction. 68. Volume of a Solid of Revolution. The area bounded by the curve y = f{x), Fig. 46, the ordinates x = a and x = b, and the X-axis, is revolved about the X-axis. Find the volume of the solid generated. Divide the interval AB = 6 — a on the X-axis into n equal parts of length Ax and pass planes through the points of division perpendicular to the X-axis. These planes divide the volume into H 110 CALCULUS [§68 n portions, Ay. A typical portion can be regarded as generated by revolving DEQP, Fig. 46, about the base DE in the X-axis. Replace the volume of this slice by that of the cylinder generated by the revolution of DEQ'P about the Z-axis. Its volume is TTj/'Ax. The total volume is then or -r. Illustration. Find the volume between the planes x = 1 and X = 3 of the solid generated by revolving the curve y = x^ -\- x about the X-axis. /»3 /»3 F = TT I {X^ + Xydx = TT \ {x* + 2x3 _|_ x^)(lX = X[U' -I- \X* -I- W]\ = TTXV^X^ + hx + ii; = 27ir(| -1-3-^1)- x(i + i 4- i) = -^ tPr. Exercises 1. Find the volume between the planes x = and x = 3 of the solid generated by revolving the parabola y^ = 6x about the X-axis. 2. Find the volume of a sphere of radius r. 3. Find the volume of the ellipsoid of revolution generated by revolving the ellipse 16 "^ 9 about the X-axis; about the K-axis. 4. Find.the volume between the planes x = and x = 4 of the solid generated by revolving y^ = x^ about the X-axis. 6. Find the volume of the solid generated by revolving x' + y^ = a' about the X-axis. 6. Find the volume generated by revolving y^ = 2ax — x* about the X-axis. 7. Find the volume generated by revolving the oval of y* = x(x — l)(x — 2) about the X-axis. §69] DIFFERENTIALS 111 69. Length of Arc: Rectangular Coordinates. In §63 the length of arc of a curve was found by integrating its differential. We shall now express the length of arc by means of a definite integral. To find the length of arc APQB, Fig. 49, divide CH into n equal parts of length Ax each. At the points of division erect ordinates dividing the arc AB into n parts of which PQ is one. The length of arc AB is defined by g ^ lim V ^ n=oo ^^ 3 P Q A / y ^ AC \ B \ O C D E ■«-o-*j H- Fig. 49. where Ac is the length of the chord PQ. Then s = ^^^ y,V(Ax)^+(Ayy Since = 1, it follows by Duhamel's Theorem that b lim s =:iT.x^^uM'^ 112 CALCULUS [§70 Hence, « = I \/l + l^) dx. r>/^^ Exercises l.^Find the length of the curve y = x^ between the points (0, 0) and'(l, 1). z 1 1 2. Find the entire length ofx* +y* = a^- 3. Find the entire length of x^ + y^ = a^. 4. Find the length of y^ = 4:X^ between the points (0, 0) and (4, 16). 70. Area of a Surface of Revolution. The portion AB, Fig. 49, of the curve y = f{x), between the ordinates x = a and x = b, is revolved about the X-axis. Find the area, S, of the surface generated. Pass planes as in §69 perpendicular to the X-axis through the equidistant points of division of the interval CH = h — a. Denote the convex surface of the frustum of the cone generated by the revolution of DEPQ by AF. The area, S, of the surface of re- volution will be defined as the limit of the sum of the convex surfaces, AF, of these frusta as n becomes infinite, i.e., as Ax approaches zero. Then, „ lim V A p lim V o ^ + ('^ + ^V^ Ac Ay By Duhamel's Theorem we can replace y + -„" by y, since lira 2/ + |Aj/ Ax-o V, = 1- Hence, Since lim , — , \x=0 I T — = 1> 'dy\i §71] DIFFERENTIALS 113 Therefore a yds, x — a 5 = 2x I v\il + [~\''dx = 27r where ds is the differential of the length of arc. The latter form is easily remembered since 2Tryds is the area of the strip of surface generated by revolving ds, the differential of arc, about the X-axis at a distance y from it. If it is more convenient to integrate with respect to y, ds can be replaced by v^d)^"- and the limits are the values of y corresponding to x = a and x = b. Thus Wl + U) dy = 2ir\ yds. Exercises 1. Find the surface between the planes x = and x = 5 of the paraboloid of revolution obtained by revolving t/'* = 4x about the X-axis. 2. Find the surface of the sphere generated by revolving x'' + 2/^ =^* about the X-axis. 3. Find the surface of the right circular cone whose altitude is 10 feet and the radius of whose base is 5 feet. 4. Find the surface of the solid generated by the revolution of 2 2 2 a^ + 2/' = a* about the X-axis. 71. Element of Integration. The first step in setting up a definite integral is to break up the area, volume, work, length, or whatever it is desired to calculate, into convenient parts which are infinitesimals as their number approaches infinity. These parts are then replaced by other infinitesimals of the typical 8 114 CALCULUS [§72 form fix)dx, which must be so chosen that the limit of the ratio of each infinitesimal of the second set to the corresponding infinitesimal of the first set is one. fix)dx is called the "element" of the integral or of the quantity which the integral represents. Thus the element of volume is wy^dx, that of area is ydx, that of work is Fdx. If the magnitude which it is desired to calculate is broken up into suitable parts, the expressions for the elements can be written down at once. The best way of retaining in mind the formulas of §§68, 69, and 70 is to understand thoroughly how the elements are chosen. The process of writing down the element of integra- tion at once becomes almost an intuitive one. 72. Water Pressure. The pressure at any given point in a liquid at rest is equal in all directions. The pressure per unit area at a given depth is equal to the pressure on a horizontal surface of unit area at that depth, i.e., to the weight of the column of liquid sup- ported by this surface. This weight is proportional to the depth. Hence the pressure at a depth x below the surface of the liquid is given by the formula p = kx. If the liquid is water and the depth x is expressed in feet, k = 62.5 pounds per cubic foot. The method to be used in finding the water pressure on any vertical surface is illustrated in the solution of the following problems: 1. Find the pressure on one side of a gate in the shape of an isosceles triangle whose base is 6 feet and whose altitude is 5 feet, if it is immersed vertically in water with its vertex down and its base 4 feet below the surface of the water. Take the origin at the vertex of the triangle, the axis of x vertical, and the axis of y horizontal, as in Fig. 50. The altitude is sup- posed to be divided into n equal parts and through the points of division horizontal lines are supposed to be drawn dividing the surface into strips. The trapezoid KHMN = AA is a typical §72] DIFFERENTIALS 115 strip. Denote the pressure on this strip by AP. The abscissa of the lower edge of the strip is x and the pressure at this lower edge is A;(9 — x). Then the total pressure is P = n^^Xm-x)LA. (1) In accordance with Duhamel's Theorem we can replace AA by 2i/Ax. x = or Since = 2A; I (9 - x)ydx. (3) V = 3x V •5 P = ~ \ (,9-x)xdx _6kr 5 Jo = g- "2- - 3^ = 5312.5 pounds. (4) In general, if u denotes the depth below the surface of the liquid and z denotes the width, at the depth u, of the vertical surface on which the pressure is to be computed, P = k \ uzdu, (5) where a and b are the depths of the highest and lowest points, respectively, of the surface. For, p = lim X\kuAA = lin^ X kuzAu = k \ uz du. n^a, ^^ Au=0 •^T I 2. Find the total pressure on a vertical semi-elliptical gate whose major axis lies in the surface of the water, given that the semi-axes of the ellipse are 8 feet and 6 feet. Take the origin at 116 CALCULUS [§73 the center, the axis of x horizontal and the axis of y positive down- ward. The element of pressure is and the total pressure is 2kyx dy ^ = 4 yxdy. X is expressed in terms of y by means of the equation of the ellipse, Then — -u ?^ = 1 64 "^ 36 P = 2ki I yVSQ - y^ dy. Exercises 1. Find the pressure on the vertical parabolic gate, Fig. 51: (o) if the edge AB lies in the surface of the water; (b) if the edge AB lies 5 feet below the surface. 2. Find the pressure on a vertical semicircular gate whose diameter, 10 feet long, lies in the surface of the water. 73. Arithmetic Mean. The arithmetic mean, ^, of a series of n numbers, Oi, 02, aa, • • • , a„, is defined by the equation or nA = ai + a2 + as + • • • + a„, ai + a2 + as + • • • + a„ A = That is, A is such a number that if each number in the sum §741 DIFFERENTIALS 117 Oi + c[2 + cfs + * • ' -\- ttn be replaced by it, this sum is unaltered. 74. Mean Value of a Function. We can extend the idea in- volved in the arithmetic mean to other problems. Illustration 1. Suppose a body moves with uniform velocity a distance of 1 foot during the first second, a distance of 2 feet during the second second, a distance of 3 feet during the third second, and so on for 10 seconds. At the end of 10 seconds the body would have moved 1 + 2 + 3+ • • -+10 = 55 feet. The mean, or average, velocity of the body is the constant velocity with which the body would describe this distance in the same time. It is equal to 5.5 feet per second. If the velocity of the body instead of changing abruptly as indicated above were changing continuously in accordance with the law V = t, the total distance s traversed in 10 seconds would be •10 tdt = 50. J "10 /*1 vdt = j Jo From this equation The mean velocity, V, the constant velocity which a body must have in order to traverse the same distance in the same length of time, is 50 -i- 10 = 5 feet per second. This can be expressed by the formula J "10 /»10 Vdt = I vdt. Jo vdt 10 In general, if i; = j{t), the mean velocity, V, of the body in the interval of time between t = a and f = 6 is expressed by the equation = { At)dt, or, since F is a constant, S{t)dt V = Vdt= I a Ja Ja b — a 118 CALCULUS [§74 V is the constant velocity, which replacing the variable velocity, " = /(O, at every instant in the interval between t = a and t = b, gives the same distance traversed, i.e., leaves the value of the nb integral, I v dt, unchanged. Illustration 2. Consider the work done by a variable force / acting in a straight hne, the X-axis, and producing a displacement from a; = a to X = 6. If the law of the force is f = (f){x), the mean force F in the interval from a; = a to x = 6, or the constant force which would do the same work while producing the dis- placement 6 — o, is given by the equation nb f*b I Fdx = I 0(x)dx, f. or (l){x)dx F = b — a F is a constant such that if, in the integral I 4>{x)dx, the func- tion ^(x) be replaced by it, the value of the integral remains unchanged. Illustration 3. Let a unit of mass be situated at each of the points on the X-axis whose abscissas are Xi, X2, X3, • • •, x„. The X-axis is taken horizontal and the masses are acted upon by gravity. We shall find the distance, x, from the origin at which the n masses must be concentrated in order that the sum of the moment about the origin of the forces acting on the masses shall be unchanged. Clearly x must satisfy the equation gnx = g{xi + X2 + X3 -{- • • • + x„), or Xi + X24-X3+ • • . +x„ X = • n If there are mi, mj, ms, • « • , 7n„ units of mass concentrated at Xi, X2, Xs, • • • , x„, respectively, the mean moment arm, x, the distance from the origin at which the masses must be con- §74] DIFFERENTIALS 119 centrated in order that the sum of the moments about the origin of the forces acting on the masses shall be unchanged, is given by the equation (mi + m2 + • • • m„)x = niiXi + W2X2 + • • • + m„x„, or - 2 THiXi 2wi -, (i= 1,2,3,. . .,n). (1) rr is a constant such that if in the sum 2 m,x, each of the num- bers Xi, X2, • • • , x„ be replaced by x this sum is not changed. Now let there be a continuous distribution of matter along the X-axis from x = a to x = b. Divide the interval b — a into n segments each of length Ax. An expression for the approximate sum of the moments about the origin of the forces acting on the mass is 2 gxAmt, where Am, is the mass of the segment AZf. An expression for the approximate force is 2 ^Am,-. Hence an expression for the approximate x is „ . — -*• It is readily seen that as Ax approaches zero, the numerator approaches the total moment and the denominator approaches the total mass. Hence lim ^ ^ T^ Am=o ZigAm I dm (2) X is a constant such that if in the integral, I x dm, x is replaced , I xdm, by X, the value of the integral is unchanged. For example, if the density is proportional to x^, i.e., is equal to kx^, the element of mass, dm, is kxHx, and we have J'*b nb xkxHx I x^dx _ a _ Ja •6 Ch Mx 3 b* - a\ »b 7'6 4 fe» — a' kx^dx I x^dx a Ja The mean value, M, cf the function fix) with respect to the 120 CALCULUS [§74 magnitvde u, which is a function of x, is defined by the equation (3) where M is a constant, or Jf*x = b r*x = b Mdu = I f{x)du, x = o Jx = a J'*x = b f{x)du x = a M = '1 = 6 du J^di x = a (4) M is a constant such that if the function f{x) is replaced by it in Jr*x = b f{x)du, the valv£ of the integral is not changed. a; = o _ In (2), X is the mean value of x with respect to the magnitude, m. A particular case of (4) is that in which u = x. Then (4) becomes M = ,— ^^ I f (x) dx. (5) 1 p Illustrations 1 and 2 are cases of this type. When w = a;, as in equation (5), M can be interpreted as the altitude, AC, of a rectangle wth base AB = b — a, Fig. 52, whose area is equal to the area bounded by the curve y = ■X f{x), the A'-axis, and the ordi- nates x = a and x = b. From this standpoint M is called the mean ordinate of the curve y = f{x) in the interval from x = a to x = b. Illustration 4. Find the mean ordinate of the curve y = x' between the ordinates a; = and x = 2. Fig. 52. M = ^-^ I f{x)dx = § I x^dx = ix3 4. §74] DIFFERENTIALS 121 Illustration 5. Find the mean with respect to w of x between the limits u = \ and tt = 9, if m = x^. J'*u =» 9 (*u = 9 Mdu = I xdu, u = 1 Ju = 1 J 3 /»3 2xdx= I 2x»dla:, 8M = V, M = V. Exercises Find the mean ordinates for the following curves: 1. y = X* between x = and x = 3. 2. 2/ = x^ between x = 2 and x = 4. 3. 2/ = 3x3 between x = and x = 2. 4. y = 3x3 between x = 1 and x = 3. 5. J/ = x* between x = and x = 1. 6. Find the radius of the right circular cylinder of altitude 3 whose volume is equal to the volume between the planes x = 2 and x = 5 of the solid generated by revolving 7/ = x + x'' about the X-axis. 7. Find the radius of the right circular cylinder of altitude h — a whose volume is equal to the volume between the planes x = a and X = fe of the solid generated by revolving y = /(x) about the X-axis. 8. The density cf a thin straight rod 10 inches long and of uniform cross section is proportional to the distance from one end. Find the mean density of the rod. 9. Find the mean velocity of a freely falling body between the time ^ = 1 second and t = 3 seconds. 10. The density of a rod is given by p = 3x2, 'v^here x is the distance from one end. Find the mean density if the rod is 10 inches long. 11. Find the mean moment arm in the case of the rods of Exercises 8 and 10, about a horizontal axis through the end of the rod (x = 0). The rods are horizontal, and perpendicular to the axis about which moments are taken. The rods are supposed to be acted upon by forces due to gravity alone. 12. Find the mean ordinate of a semicircle, the ends of which are upon the X-axis. CHAPTER VIII CIRCULAR FUNCTIONS. INVERSE CIRCULAR FUNCTIONS Up to this point only functions have been discussed which are simple algebraic combinations of powers of the dependent variable. Many interesting applications of the calculus to the study of these functions have been given. We shall now take up the study of the appUcation of the methods of the calculus to another very important class of functions, the circular functions. It is apparent that the principles developed in the preceding chapters are equally applicable to the circular functions and to the algebraic functions. As the student has already learned, the circular functions occur very frequently in the study of the physical sciences and their applications, because by means of them periodic phenomena can be studied. 76. Derivative of sin u. Let y = sin u. y -\- Ay = sin (u + Am), Ay = sin {u + Aw) — sin u — sin u cos Au + cos u sin Au — sin u, Ay _ cos w sin Am sinM(l — cosAm) Aw Am Am Then lim A^ _ lim sin Am . lim 1 - cos Am A«=o Am ~ ^°^ "^"=0 Am ®^° ^^«-o Am * Hence by §56 and §58 dy ^ = cosw.. (1) Whence dy du di = «°«^dx" (2) 122 §75] CIRCULAR FUNCTIONS 123 The corresponding formula for dy is dy = cos udu. It has thus been shown that d(smu) du — J = cos u j- dx dx (3) (4) and d(sinu) = cos udu. Well known properties of the function y = sin u can be verified by formula (1). Thus sin u is an increasing function between tt = and u = t), and between u — -^ and u = 27r, and decreas- ing between w = ^ and w = -p" The same facts are shown by re I I I p^ I I I I I I I I I I I G_u Fig. 53. the derivative, cos u, which is positive between w = and « = o' Stt and between u = -^ and u = 27r, and negative between u = ^ and u = Stt Further, sin u has maximum and minimum values for u = ^ and u = -^» respectively. The same facts are shown by the derivative, cos u, which becomes zero at these points and changes sign at ^ from plus to minus, and at -^ from minus to plus. The slope of the sine curve is approximately the slope of the diagonal PQ of a rectangle in Fig. 53. The greater the number of equal parts into which the circumference of the circle is divided and hence the smaller the subdivisions of the arc, the closer do the slopes of these diagonals approach the slopes of the tangents. 124 CALCULUS [§76 76. Derivatives of cos u, tan u, cot u, sec u, esc u. The derivatives of the remaining circular functions can be obtained from that of the sine. Let y = cos u. Then y = sin and dy /x X^(I-^) ^ = cos^2-i.j- ^^^ du\ = cos (I - u) (- ^) du = — sm « -r' dx Hence d(cos u) . du — ^ = — sm u :i— dz dz and d(cos u) = — sin u du. By writing sin u tan u = » cos u cos u cot u = --. » sin u and sec u 1 CSC u = (1) sin u the student will show that d(tan u) du — ^ = sec^'u^ » or d(tan u) = sec^u du (2) d(C0t U) du Ar 4. \ 2 A f}\ — J = — csc'^u , » or d(cotu) = — csc^ u du (3) d(sec u) * du ^r \ *. a /^^ — -V = sec u tan u , ' or d(sec u) = sec u tan u du (4) d(cscu) . du ,. . , , ._. - — J = — CSC u cot u J » or d(csc u) = — csc u cot u du (5) §76] CIRCULAR FUNCTIONS 125 Illustration 1. Find the first and second derivatives of 3 sin (2a; - 5). rf[3sin (2a; -5)] _ d[sm (2x - 5)] dx dx d(2x - 5) dx = 3 cos (2a; - 5) = 6 cos (2a; — 5). Differentiating again, d'[3sin (2a; -5)] ^ rf[cos(2x-5)] dx^ dx = —6 sin (2x — 5) ^ = -12sin(2x - 5). dv Illustration 2. li y = sin 2a; cos x, find j— Since sin 2x cos x is the product of two functions, apply formula (1) §40. dy • „ / . ^ dx . , X / ^ X d2x -T- = sin 2a;( — sin x)-i — |- (cos x) (cos 2x) -r- = 2 cos x cos 2a; — sin x sin 2x. dv d^v Illustration 3. If ?/ = 3 sin x + 4 cos x, find --r- and -r-^' -p = 3 cos X — 4 sin X (6) From (6) d^y J— 2 = —(3 sin X + 4 cos x) = —y. (7) -r- = 4 cos X (f — tan x). TT When < x < ^, cos x is positive. The second factor, f — tan x, is positive when x < tan~^ (^), and negative when x > tan"^ (f ). Thus, when x is in the first quadrant the function has a maximum value coriiesponding to x = tan~i (|). When r, < a; < TT, -T- is negative. 3x When TT < X < -^, cos x is negative, and J — tan x is negative 126 CALCULUS [§76 when X < tan~^ (f), and positive when x > tan-' (f). Thus when X is in the third quadrant the function has a minimum value corresponding to the value x = tan~^ (j). When -^ < x< lir, -3- is positive. The same facts can be seen directly from the function, for it can be put in the form y = 5(5 cos X + t sin x). Let cos a = i and sin a Then or y = 5(cos x cos a + sin x sin a), y = 5 cos (x — a). In polar coordinates this represents a circle passing through the origin, with a diameter of 5. (See Fig. 54.) x is the vectorial angle and y the radius vector. The diameter OB makes an angle a with the polar axis. As x varies from to IT the circle is described, and as x varies from tt to 27r, y is negative and the circle is described a second time. Hence y has a maximum value 5 when X is equal to a, and a mini- FiG. 54. mum value —5 when x is equal to a -\- IT. dy d^y Illustration 4. li y = tan' 3x = (tan ZxY, find -r- and -j-^* The function is of the form y = w. Hence g = 3(tan3x)^^^^^ = 3 tan^ 3x sec^ 3x d3x dx = 9 tan^ Zx sec^ 3x. d^ dx* 9 tan* 3x c?(sec* 3x) dx + sec* 3x (i(tan* 3x)- dx §76] CIRCULAR FUNCTIONS 127 = 18 r tan2 3x-2 sec 3a; ^ , ' + sec^ 3x-2 tan 3x - ■ dx dx dSx tan^ 3a; sec 3a; sec 3a; tan 3a; , dx dSx^ + sec'^ 3x tan 3a; sec^ 3x -i — = 54 (tan' 3x sec^ 3a; + tan 3x sec* 3a;) = 54 tan 3a; sec'' 3x(tan2 Sx + sec' 32). Illustration 5. If -r- = cos x, find y. dy -T- = COS x. dx y = sinx -\- C. dv Illustration 6. If -p = cos 3a;, find y. dy .r „ d3a;"| ^=i[cos3x^J. The expression within the bracket is the derivative of sin 3x, hence, 2/ = ^ sin 3x + C. dv Illustration 7. If ",^ = sin 3a;, find y. Hence = -i[-sin3a;^]. ?/ = — ^ cos 3a; + C. Illustration 8. If -j- = sec^ 2x, find y. 1/ = 5 tan 2a; + C. dv Illustration 9. If -^- = sec 5x tan 5x, find y. Hence = I sec 5x tan 5x -r— . 128 CALCULUS [§76 Hence y = 6 sec 5a; + C. Illustration 10. li dy = cos 3x dx, find y. y = \ cos Zxdx = \ \ cos 3x rf(3x) = \ sin 3x + C'. Illustration 11. li dy = sin'^ 2x cos 2x dx, find ?/. y = j (sin 2x)^ cos 2x dx = i 5/3 (sin 2x)2 cos 2x rf(2x) = i J3(sin 2xyd{sm 2x). Hence 2/ = Ksin2x)« + C. Illustration 12. If d?/ = tan^ 5x sec^ 5x dx, find y. y = j tan^ 5x sec^ 5x dx = i i J 4 (tan 5x)^ sec'' 5x d(5x) = "sHr j 4 (tan 5x)' d(tan 5x). Hence y = ■2V(tan Sx)^ + C. Illustration 13. J sin 5x cos 3x dx = J Msin (5x + 3x) + sin (5x — 3x)]dx = 5 I sin 8x dx + ^ J sin 2x dx = — iV cos 8x — J cos 2x + C. Illustration 14. j cos 7x sin 3x dx = J |[sin (3x + 7x) + sin (3x — 7x)]dx = I J sin lOx dx — 5 J sin 4x dx = — 2V cos lOx + I cos 4x + C. §76] CIRCULAR FUNCTIONS 129 Illustration 15. J cos 4a; cos 7x dx = J | [cos (7x + 4a;) + cos (7x — 4a;)] dx = 5 J cos llx dx + hi cos 3a; dx + C = -2V sin 11a; + 6 sin 3a; + C. Illustration 16. J sin 4x sin 2x dx = — ^ J [cos (4a; + 2x) — cos (4a; — 2x)] dx — ~ 2] cos 6x dx -{- h j cos 2a; rfx = — iV sin 6x + J sin 2x + C. Exercises In Exercises 1 to 10, verify the differentiation. 1. y = sin ox, ^ = 5 cos ox, -3-^ = — 2oy. dy „ . „ d^y 2. y = cos 3a;, dx ^ ~ ®'" ' dx^ " ~ ^^* 3. y = tan 2x, 'd ^ '^ ^^^^ ^^' j-^ = 8 sec'' 2x tan 2x. dw 4. 2/ = sin X cos 2x, -r- = cos 2x cos x — 2 sin 2x sm x . 3x - 2 dy , 3x - 2 6. 7/ = sin — ^ — = I cos dx ' 5 ' dx-^ = - ^' y- 6. 2/ = tan' 5x, dy = 15 tan^ 5x sec* 5x dx. 7. y = sec^ 3x, dy = 12 sec* 3x tan 3x dx. [y ~ °(1 ~ cos 0), dy = a sin d9. Lx = a{d — sin 0), dx = a(l — cos O)d0. . /2wt \ dy 2aw /27rt \ 9. y = a sin I ^^ ^) ' di ^ ^ ^°^ XT' ~ V ' 10. y = X sin x, dy = (x cos x + sin x) dx. 11. From the results of Exercise 8, show that -r- = cot s- ' dx 2 130 CALCULUS [§76 Find dy in Exercises 12-20. 12. y = tan 2xsin 2x. 15. y = cos (3 — x)*. sin 2x 14. y = sin (x^ + 3x — 2). 17. y = x cos 2x — tan 2x. 18. y = tan- (x — 1). 19. y = cos* (1 — x» — 2x). 20. 2/ = sin2 (2x - 1) cos^ (2x - 1). Integrate : 21. dy = sin 2x<fx. 26. dy = sin x cos xdx. 22. dy — cos 2xdx. 26. dy = tan x sec* xdx. 23. dy = sec* 4x<ix. 27. dy = Vsin 2x cos 2xdx. 24. dy = sec 5x tan 5xdx. 28. dy = cos^x sin xdx. 29. dy = sec* x tan xdx = sec'x sec x tan xdx. 30. dy = sec" (x — 1) tan (x — 1) dx. 31. Find the area iinder one arch of the sine curve. 32. Find the area under one arch of the curve y = 2a* sin* x. 1 — cos 2x Hint, sin* x = „ 33. The equations of Exercise 8 are the parametric equations of the cycloid. Find the length of one arch of the cycloid. Hint, ds — -\/{dxy + (dy)". Express ds in terms of e and dO. 34. Find the area under one arch of the cycloid. dv 36. X = a cos 6, y = a sin 0. Find -j-. Find the length of the curve. Find the area bounded by the curve 36. x = the curve. dy 37. X = a cos' <j), y = a sin' </>. Find -j-. Find the length of the curve. 38. Find the volume bounded by the surface obtained by revolv- ing y = sin X about the X-axis. 39. A man walks at the constant rate of 4 feet per second along the diameter of a semicircular courtyard whose radius is 50 feet. The sun's rays are perpendicular to the diameter. How fast is the man's shadow moving along the semicircular wall of the courtyard when he is 30 feet from the end of the diameter? dv 36. X = a cos e, y = h sin 6. Find t-. Find the area bounded by §77] CIRCULAR FUNCTIONS 131 40. A drawbridge 25 feet long is raised by chains attached to the end of the bridge and passing over a pulley 25 feet above the hinge of the bridge. The chain is being drawn in at the rate of 6 feet a minute. Horizontal rays of light fall on the bridge and it casts a shadow on a vertical wall. How fast is the shadow moving up the wall when 13 feet of the chain have been drawn in? 41. Find ^ ii x = y\/y - 1. dy .^ 42. Find ^ ii x = Vl - sin y. 43. If p' = a* cos 2d . show by implicit differentiation that dp _ a* sin 29 dd^ P 44. If p 2 cos = a^ sin 36, find ^• 46. I sin 6x cos 2x dx. 49. I sin Ax cos 7x dx. 46. I cos 4a; cos 3x dx. 60. I cos 5x cos 9x dx. 47. I cos 5x sin 2a; dx. 61. I sin wt cos at dl. 48. I sin Sx sin 3x dx. 62. 1 cos ut cos at dt. 63. Find the mean ordinate of the curve y = sin x between the limits X = and x = tt. 77. Derivatives of the Inverse Circular Functions. ^ The for- mulas for the derivatives of the inverse circular functions are readily obtained from those of §§75 and 76. . 1 The student will recall that sin"' u is defined for values of u between —1 and +1 only, and that it is a many valued function. To a given value of u there correspond infinitely many angles whose sines are equal to u. This will be seen to be the case on sketching the curve y = sin"' u. In this and future discussions of this function it will be made single valued by considering only those values of y = sin"' m which lie between — ^ and +o'> inclusive. The positive sign of the radical in the final formula (1) is chosen because cos V ■» \/l — uJ is positive when j/ lies between —^ and +0"* Of the functions occurring in (2), (3), (4), (5), and (6), y = cos"' u, and y = sec"' u are made single valued by choosing y between and ir, while the remaining functions, y — tan"' u, y =• cot"' u, y = esc"' u, are made single valued by choosing y between — 2" and +2"' Show that the proper sign has been chosen for the radicals in the formulas (2), (5), and (6). 132 CALCULUS [§77 Let y = sin"^ u. Then sin y = u. Differentiation gives dy du cos y -r = T~' ^ dx dx dy _ 1 du dx ~ cos y dx 1 du Vl — sin" y dx du dy dx Hence dx y'l _ u^ Therefore, du d(sin-iu) dx J/ . , X <*" (1) — -— = , » or d(sm-i u) = —7 - '^ ' The student will show that du d(cos-iu) dx J/ , X du (2) : = — , > or dtcos-^u) = ,■ ^ ' dx . VI - u2 VI - u' du d(tan-»u) dx ^/* , x <^^ (3) or d(tan-'u) = f^p^z du or d(cot-i u) = - YJ^i (4) » or d(sec"^u) = 7 - (5) uVu* - 1 - , ^ or d(csc-i u) = , (6) dx uVu" - 1 uVu^ - 1 Illustration 1. K y = sin"i(x^ — 2x — 3), find dy. By- formula (1) dx " 1 + u^' du d(cot-» u) dx dx " i + u*' du d(sec~^ u) dx dx u Vu2 - i du d(csc-' u) dx §77] CIRCULAR FUNCTIONS 133 d(x^ -2x-3) dy = Vl - {x^ -2x- 3)« 2(x - l)dx Vl - (x* -2x- sy Illustration 2. U y = tan~^ 3x, find dy. By formula (3) _ djSx) ^y ~ 1 + (3x)2 _ 3(Za; ~ 1 + 9x2" Illustration 3. If dt/ = 7-3^ — ^» find y. /r (ia; + a;2 or y = tan~^ x + C. Illustration 4. li dy = ^ , ^ ^ y find r/. cix ■'/r + 9x» 3(ix + (3x)2 The expression under the integral sign is now of the form -r-y — i whose integral is tan~^ u. Hence y = \ tan-i (3x) + C. (XX Illustration 5. If dy = . . „ g > find ^. -/ dx 4 + 9x2 dx + {ixy 134 • CALCULUS Hence y = i tan-i (ix) + C. Illustration 6. If dy = — ^ , find « V4 - 9x2 »• [§77 9x5 _ j^ r dx ~ V Vl-(|x)»' = 1.2 r ^dx ' VVi-(ix) The expression under the integral sign is now of the form du ;^/p=| whose integral is sin-» u. Hence y = i sin-i (fx) + C. Exercises dy Find ^ in Exercises 1-10. 1. y = sin-i (x^). 6. 2/ = sin"' (sin x). 2. y = sin-i (x - 1). f, y = tan-i -^— • X — 1 3. 2/ = tan-Ux2). 8. t/ = sin-» (1 - x)K 4. y = tan-i (x - 1). 9. y = gec-i (a;2 _ 3). 5. y = sm (sin-»x). 10. y = a;sin-ix. Integrate : 11. dy = t4%- ^ 4 + x2 12. rfw = K-^^- ^ 9 + x2 13. rfy = "^^ 25 + 16x2 -^ 7 _ dx It ^ ~ a2 + x^' ^^' y = a *^"~' ^ + ^- CIRCULAR FUNCTIONS 135 dx Ans. y = sin~* — f- C dx 20. dy = . — = -• Ans. y = - sec~i - + C. Using the results of Exercises 14, 18, and 20 as formulas, evaluate the following integrals: 21. 22. 23. 24. 26. 26. 27. J x/16^^^" ^^' J J Vl6 - 9x» J ^ J Vl - 9x2 J J 25TT^ 32. J - J 25 + 16x2* 33. J dx x\/9x2 - 1 dx X2 + 17 dx Vl3 - a;2 dx \/x2 - 19' dx 5x2 + 8 dx J 1 + 16x2" 34- J ^2^43.4.5 J J 7v^2^^- 35. J ^ (x+2)2+l dx x\/3x2 - 14 C dx «„ r dx 28. I -. 36. I , J x\/9x2 - 25 J \/9 - x2 37. Find the area between the ordinates x = 0, x = §, the X-axis and the curve y = — . Vl -x2 4a' 38. Find the area under the curve y = , , . , » above the X-axis, " x2 _j. 4o2 ' 136 CALCULUS [§78 and between the ordinates x = and x = b. Find the limit of the area as b increases without limit. 39. Find the mean ordinate of the curve y = j--r — ^> between the ■JT limits X = and x = -y 78. Velocity and Acceleration. If a particle is moving in a curved path, its velocity at any point is represented by a vector laid off along the tangent with its length equal to the magnitude of ds the velocity, -r:' Thus the ve- locity at the point P, Fig. 55, is represented by the vector PT. It can be resolved into the com- ponents PK and PM, parallel to the X- and F-axes, respectively. ~^ These components represent the time rates of change of the coor- dinates of the moving point P, i.e., PK^^"" Fig. 55. and Since PM = dt dy dt' (1> PT = V{PK)^ + {PM)\ ds dt (2) This relation can be obtained directly from (2), §63, if we consider X and y functions of t. For, we can divide by dt and obtain the equation (2). In Fig. 55, let PT be the velocity at P, and QT' that at Q. Draw from a common origin, o, Fig. 56, the vectors op and oq equal to the vectors PT and QT', respectively. Then pq equals the vector increment, Av. The average acceleration for the Av interval At is equal to -rr directed along pq. Lay off, on pq, pm §78] CIRCULAR FUNCTIONS 137 equal to ry As At approaches zero, Q approaches P, and q ap» proaches p as indicated by the dotted line, Fig. 56; p7n approaches a vector pt directed along the tangent to the arc pq at p. This vector, Av the limit of — .• represents the acceleration of the particle moving in the curved path. Let us calculate its x and y components. In Fig. 56, denote: op by V and its components by Vx and Vy, oq by v' and its components by v'x and v'y, pq by Ay and its components by Ay, = v'x — Vx and Ay„ = v'„ — Vy, pt by j and its components by jx and j^. Fig. 56. Then Jx = lim Ay. _ dv^ ^'=^0 Af dt lim Ay^ dv„ dt dt dt d^ df d^ dt"^' The magnitude and direction of the vector ^ are given by: (3) (4) (5) 138 CALCULUS [§79 and where is the angle made by pt, Fig, 56, with the positive direction of the X-axis. Again we can resolve the acceleration j into components along the tangent and normal. In Fig. 57, PL is the tangential com- ponent and PJ is the normal component. The tangential com- ponent clearly produces the change in the magnitude of the velocity, and the normal component the change in its direction. Fig. 57. 79. Angular Velocity and Acceleration. If a body is rotating about an axis, the amount of rotation is given by the angle d through which a line in the body turns which intersects the axis and is perpendicular to it. Thus in the case of a wheel the rotation is measured by the angle 6 through which a spoke turns. ^ is a function of the time t. The rotation is uniform if the body rotates through equal angles in equal intervals of time. If the uniform rate of rotation is co radians per second, the body rotates through 6 = cat radians in t seconds. If the rotation is not uniform the rate at which the body is rotating at any instant, the angular velocity, is <^^M^o^t dt' §791 CIRCULAR FUNCTIONS 139 Similarly, the angular acceleration a is the time rate of change of the angular velocity. Then, do) dW "~lt ~ dt^' If we consider a particle at a distance r from the axis of rotation, its linear velocity v is V = cor and is directed along the tangent to the circle described by the particle. The tangential acceleration is jt = ar. Exercises , 1. The following formulas have been established for linear motion, with constant acceleration : V = Vo + jt. S = Vot + ^jl\ I' - ^* = js. (See §38.) Show that the corresponding formulas for rotation are: e = cjoi + W^' W^ COo^ T - "2- = «^- 2. A flywheel 10 feet in diameter makes 25 revolutions a minute. What is the linear velocity of a point on the rim? 3. Find the constant acceleration, such as the retardation caused by a brake, which would bring this wheel to rest in 30 seconds. How many revolutions would it make before coming to rest? 4. A resistance retards the motion of a wheel at the rate of 0.5 radian per second per second. If the wheel is running at the rate of 10 revolutions a second when the resistance begins to act, how many revolutions will it make before stopping? 5. A wheel of radius r is rotating with the uniform angular velocity w. Find the direction and magnitude of the acceleration of a point on the rim. 140 CALCULUS [§80 Hint. The coordinates of the point can be written x = r cos w<, y = rain. uL Find -jtj and tt^* 6. A wheel of radius r is rolling with the uniform angular velocity u along a horizontal surface without slipping. How fast is the axle moving forward? The parametric equations of a point P on the rim are: X = r{u}t — sin cat) y = r(l — cos (at). Find the magnitude and the direction of the velocity of P at any instant. What is the velocity of a point at the top of the wheel? At the bottom? 7. If a particle moves in such a way that its coordinates are re = a cos t -\- h, y = a sin < + c, where t denotes time, find the equation of the path and show that the par- j> tide moves with constant tangential velocity. 80. Simple Harmonic Motion. Let the point P, Fig. 58, move upon the circumfer- ence of a circle of radius a feet with the uniform velocity of v feet per second, so that V the radius OP rotates at the rate of - = co Fig. 58. o, radians per second. The projection, B, of P on the vertical diameter moves up and down. If the point P was at C when t = 0, the displacement, OB = y, is given by y = asind = asinwf. If the point P was at D when t = 0, we have y = asin(co^ — a). (1) Any motion such that the displacement at time t is given by (1) is called a simple harmonic motion. Thus the point B, Fig. 58, describes simple harmonic motion. The abbreviation "S.H.M." will be used for "simple harmonic motion." From (1) it follows that the velocity of a point describing S.H.M. is dy dt (2) §81] CIRCULAR FUNCTIONS 141 and that the acceleration is -77^ = —a cousin (co^ — a). (3) The second member is —oihj, by equation (1), Hence - a)2y, (4) or df" + co^y = 0. (5) Equation (4) shows that the acceleration of a particle describing S.H.M. is proportional to the displacement and oppositely di- rected. That the acceleration is oppositely directed to the displacement is to be expected from the character of the motion, which is an oscillation about a position of equilibrium. Thus if the body is above this position the force is directed downward, and vice versa. In Fig. 58, the point B has a positive acceleration when below and a negative acceleration when above 0. The acceleration is zero at 0, a maximum at the lower end of the diame- ter, and a minimum at the upper end. In accordance with (2) the velocity is zero at the two ends of the diameter. The velocity has its greatest numerical value when B passes through in either direction. Equation (4), or (5), is called the differential equation of S.H.M. The proportionality factor co^ is connected with the period T by the relation T = — The equation (4) was obtained from (1). Frequently it is desired to solve the converse problem, viz., to find the motion of a particle whose acceleration is proportional to the displacement and oppositely directed. In other words, a relation between y and t is sought which satisfies equation (4). Clearly (1) is such a relation. However, it will be instructive to obtain this relation directly from (4). First, a differential equation equivalent to (4) will be obtained in the solution of the problem of the motion of the simple pendulum. 81. The Simple Pendulum. Let P, Fig. 59, be a position of the bob of a simple pendulum at a given instant and let it be moving to the right. If s denotes the displacement considered positive 142 CALCULUS [§81 on the right of the position of equiUbrium, -tt^ is the acceleration ds in the direction of the tangent PT, for ^ is the velocity along the tangent. This acceleration must be equal to the tangential component of the accelera- tion due to gravity, if the resistance of the air be neglected. This component is equal to — g sin 9. Since it acts in a direction opposite to that in which s is increasing, it must be taken mth the negative sign, i.e., the acceler- ation diminishes the velocity. We have then Fia. 59. d's . . (1) If the angle through which the pendulum swings is small, sin 6 can be replaced by 6. Then (1) becomes d^ dt^ Since s = W, (2) d^ dV I (3) Putting , = w^ for convenience in writing, d-'d dt^ = - co^d. dd Multiplying by 2 -ir and integrating. \dt) - oiW^ + C». (4) The arbitrary constant is written for convenience in the form C. The constant must be positive. Otherwise the velocity would be imaginary. Extracting the square root, dd dt = VC*- w2 0»' or dd Vc^ - w' d^ = dt. (5) §81] CIRCULAR FUNCTIONS 143 Integration gives — sin~^ -p7 = t -\- Ci. sin~^ -^ = cof + coCi = «< + C2, where the constant coCi is replaced by the constant C2. Then -T7 = sin (wf + C2), = — sin (a>« + C2) CO = C3 sin {wt + C2), where — has been replaced by C3. Therefore CO e = Ci sin (co< + C2) (6) is the equation of the angular displacement of the pendulum. 2_. /r The form of (6) shows that the motion is of period — = 2ir'\^-' It is a S.H.M. and contains two arbitrary constants. They can be determined by two conditions, e.g., the displacement and velocity at a given instant. Suppose the bob drawn aside to the right so that the string makes an angle 9o with the vertical. The bob is then released without being given an impulse; i.e., with an initial velocity zero. The time will be counted from the instant of release. The conditions are then d = do (7) and f = » («) when t = 6. From (6), do -jj = C0C3 cos (cof + Ci). The condition (8) gives = C0C3 cos C2, or cos C2 = 0. Whence C2 = ^- Then (6) becomes 6 = Cz cos cot. 144 CALCULUS [§81 The condition (7) gives do = Cz. Hence e = do cos oit. (9) Multiplying by I and recalling that W = s, and denoting Ido by So, we have as the equation for the displacement s, S = So cos bit. (10) The period is T = — = 27r 'y-. When is the velocity of the bob greatest? When least, numerically? Equation (6), the solution of (4), shows that, if the acceleration of a particle is proportional to its displacement and oppositely directed, the particle describes S.H.M. Exercises 1. Write the differential equations of the following simple harmonio motions. Find the period in each case. y = 5 sin 3^ 2/ = 6 sin Izt + Ij • y = 5 cos 3/. 2/ = 4 sin 2t + S cos 2t. J/ = 7 sin {8t + a). 2. Write the equation of a S.H.M. which satisfies the equations: J + 3!,-0. CHAPTER IX EXPONENTIAL AND LOGARITHMIC FUNCTIONS 82. Derivative of the Exponential and Logarithmic Functions. Let T/ = a*. (1) Then Ay = a'(a^ - 1) Ay _ ^/(i^ — 1\ Ax~ \ Ax ) dx~ ^ Ax^o Aa; ^'^> „, a^ — I . . . . , lim fl"^ — 1 . bince — -r IS independent of x, ^^q — r is a constant for a given value of a. Call this constant K, so that Then from (2), _ lim «^- 1 ,^. % = Ka: (4, Equation (4) shows that the slope of the curve y = a'' is propor- tional to the ordinate of the curve. In other words, the rate of increase of the exponential function is proportional to the function itself. When a; = 0, it follows from (2) and (3) that dy Hm a^ - 1 dx ^-0 Ax = K. Consequently the constant K introduced above is the slope of the curve t/ = a^ at the point (0, 1). This slope depends upon the value of a. Let e be that particular value of a for which the corresponding curve, y = e^, has a slope equal to 1 at the point where it crosses the F-axis. 10 145 146 CALCULUS [§82 If, then, equation (4) becomes y = 6"=, (6) dy dx since K = \'\n this case. Or de Then and dx =''• (6) de» du _. de» = cdu. (8) Equation (6) shows that the slope of the curve y = e' is equal to the ordinate of the curve. The number e is the base of the natural system of logarithms. It is sometimes called the Naperian base. Its value, 2.71828 • • • • , will be calculated later in the course. The formula for the derivative of the natural logarithm of a function is now easily obtained. In calculus if no base is indi- cated, the natural base is understood. Thus log u means log, u. If y =- log u, u = e« du dy and by (7) Whence That is or (9) dx '' = ^"d|- dy dx ' _ 1 du ~ e" dx = _ 1 du ~ u dx d(log u) _ 1 du dx ~ u dx' d(Iog 1 du u) = — • ' u §82] EXPONENTIAL AND LOGARITHMIC FUNCTIONS 147 Since ^ logoW = logoclogw, (10) d (logau) , „ ^ 1 du or (11) du d(logaU) = logaC — ' 11 y = a", log y = uloga l dy _ , du y dx ~ dx dy . du Tx^^y^^'^^'Tx . du = aMoga^. That is da" , du ^ = auloga^ da« = a«» log a du (12) Illustrations. 1. If 2/ = e*', dy = e^'d(x2) = 2xe''"dx. 2. If 2/ = e''"" ", dy = e "''^ ''d{smx) = cos x e"° * dx. r. Tr 1 / . HX 1 1 ^(^ + 1) 1 dx 3.Uy = logio (x + 1), dy = logio e ^ , ^ = logio e^-^y •l. J.1 J/ = lUJ U^ t - ^), t ''V x + V »Let z = log u Then & = u Taking logarithms to the base 1, z log a e = logo U That is log u 1 og» 1 e = logo M 148 CALCULUS [§82 5. Uy = e*^'^"'^ dy = e*'*'^'"^ d(tan-i x) = e*^"" '^ rr^ • 1 -\-x 6.1iy = log 1^^, y = 2 log (1 + x) - 3 log (1 - X), and , 2dT; , Sdx 5 + X , 7. If 2/ = e* sin x, -r- = e*(cos x + sin x) and -j-^ = 2e* cos X. Exercises Find the first derivative of each of the following : 1. y = e'*. 6. 2/ = log (1 - x^). 11. y = e^''^*. 2. y = e'\ 7. y = e' cos x. 12. y = e***^ ^'. Z. y = log (x"). 8. 2/ = e'». 13. y = log Vx'' - 1. 4. y = log (x'). 9. y = e'=^', 14. y = e~* sin x. 6. y = log (x* - 1). 10. y = log [^^^I' 16- 2/ = 10'- 16. y = logio X. 17. y = 5'. 18. y = x^ 5*. 19. Show that the aubtangent for the curve y = a'' is a constant. What is this length when a = e? Illustrations. 8. If dy = e'rfx, y = \ e'dx = e' -\- C. 9. If dy = xe'^dx. y = \ xe'^dx = |e'' + C. 10. If dy = ^> X J/ = log X + C = log X + log i2! = log Kx. §82] EXPONENTIAL AND LOGARITHMIC FUNCTIONS 149 dx 11. If dy = ^:^, y = \og{x + l) + C = log K(x + 1). xdx 12. If dy = ^^^p3, a;2 + l 2xdx x' + l = i log (x^ + 1) + (7 = log Vx2 + 1 + C = log KVx^ + 1. 13. If dy = e™* cos xdx, y = j e "° * COS X dx 1.1 Tf J (x + l)dx 14. If dy = ^2 + 2x + 3' -/ (x + l)dx x2 + 2x + 3 r 2(x + l)dx ■^ ^ J x2 + 2x + 3 = h log (x2 + 2x + 3) + (7 = log Vx^ + 2x + 3 + C = log KVx^ + 2x + 3. 15. If dy = tan x dx, y = I tan x dx -/- sin X , ax cosx = — log cos X + (7 = log sec X + C. 150 CALCULUS l§82 16. If 17. If 18. Find Let whence and Then, since dy = cot xdx, y = \ cot X dx Si dx sin X = log sin X -^ C. dy = sec x dx, y = J secx dx -\- C (sec x + tan x) sec xdx sec X + tan x /-■'■■""•" f- dx sec X -J- tan x = log (sec X + tan x) + C dx S Vx^ ± a2 v^ = x^ ± a" 2v dv = 2x dx dv X dv X dx dx V dx _ Ci dx + dv X + V ' dx + c^p 19. If Vx^ ± a2 J X + v = log (x + v) -\-C = log (a; + Vx* ± a2) + C. y ■/. e' + 1 = log (e* + 1) + C. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 151 20. If ^^ = ^, y X log 2/ = log X + log C log y = log Cx y = Cx. 21. If — = n — J y X log y = n log a; + log C = log a;» + log C = log Cx", 1/ = Cx". Exercises The results of Illustrations 15, 16, 17, and 18 are to be used as formulas of integration. In the following exercises and y : 20. dy = x2 e*' dx. 27. dy = tan 2x dx. 21. dy = e*-"" sec"^ X dx. 28. dy = cot2xrfx. 22. dy = — r-r. 29. dy = sec 2x dx. " X + 1 " 23. dy = Y^x ^°- "^^ = (logx)*^. _, - xdx «^ , cos X dx 2*- ^^ = IT^^' 31. dy = ^-j^^^^^- 26. d2/ = ^,- 32. d2/ = (!l-ZiZM£. '^ 1 — x^ -^ e* + e * «/. , (e"^ + l)dx „„ , sec^xdx (e2* - l)dx 34. dy = + 1 35. Find the area between the equilateral hyperbola xy = 10, the X-axis, and the lines x = 1 and x = 2. 36. The volume of a gas in a cylinder of cross section A expands from volume vi to volume Vi. If it expands without change in tem- perature the pressure, p, on the piston varies inversely as the volume (Boyle's Law, pv = K). Show that the work done by the expansion is pdv K\ ? = xiog^;. 152 CALCULUS [§82 37. The subtangent of a curve is of constant length, k. Find the equation of the curve. 38. For what value of x is the rate of change of logio x the same as the rate of change of z? -I I tan 36 dd. 48. | 39. I CSC X dx. (See Illustration 17.) xdx 41. j e*"'^'' sin e dd. 49. j lO'dx. 42. I cot I dd. 60. I cos^ (5x — 4) 2 (1 — Bmx)dx 43. j X sec2(z2 + l)dx. 51. | I ' I a; + cos X 44. I sec I dx. 52. j e^^ + """^ "> sin dd. Cdx C xHx 46. I — 63. I ; , 3 - 46. f'-^^. 64. f-^:^. J„ Vl+x^ J Va^ + X* 66. I (tan 20 — \)^de = \ tan 20 + log cos 20 + C. / /Idx 57. I C8c(7x + 5) dx. 68. 1 -^ +^)^" Vx^ + 4x + 7 69. ~ (3x + 2)dx "• ' 3x* + 4x + 9 §83] EXPONENTIAL AND LOGARITHMIC FUNCTIONS 153 (x + 2)dx 61 /. (x" + 4x + 7)2 62. Show that y = ae~** cos o)t satisfies the differential equation g+2fc* + („.+*=)„-0. 63. Find the mean ordinate of the curve w = - between the limits " X X = 1 and X = 2. 83. Logarithinic Differentiation. It is often advantageous in finding the derivative oi y = f{x) to take the logarithm of each member before differentiating. A number of examples will be solved to illustrate the process. (x- 1)3 lUllSLIUUUII, 1. X' lllU Liie ut {x + 1)^ y = {x - 1)^ (x+ 1)' and take the logarithm of each member. log 3 !/ = 1 log (X - 1) - 1 log (x + 1). Differentiating, Idy y dx 2 3 3(x - 1) 5(x + l) Let x + 19 15 (a;2 - 1) dy _ g + 19 dx ~ 15 (a;2 - 1) 2/ X + 19 (x - 1)^ 15 (x2- 1) (x + 1)' x + 19 15 (x - l)^ (x + 1)^ 1 11.11 RtTniinti. 2 •\/l — x^ Find the derivative of ■" i/x^ + i Vl-a;^ 2/ - 3/-^^^^ 154 CALCULUS l§83 log 2/ = § log (1 — x2) — A log (x2 4- 1) 1 dy X 2x ydx^~ 1 -x2 ~ 3 (a;2 + 1) _ _ 3^ (5 + a:") ~ 3 (1 - x^) dy a; (5 + a:') Vl - x^ dx ~ 3(1 -x^) v^mh: a: (5 + a;2) 3Vl -a;2(x2+ 1)^ This method is manifestly shorter and simpler than that of differ- entiating by the rule for the derivative of a quotient. Illustration 3. Find the derivative of {x^ + l)''''^^. y = {x"^ -\- 1)3^^2 log y = (3a; + 2) log {x^ + 1) dy J^=(^^ + 2)^-^ + 3log(x^ + l) ^^ = [(3a: + 2) ^^ + 3 log (x2 + 1)] {x^ + 1)3-+^ - -p is called the logarithmic derivative of y with respect to x. It will be considered further in a later article. Exercises Find the derivative in Exercises 1-8. 3 1. 2/ = ^^^tilL- 3. 2/ = (X + 1)3 (2x + 5)1 (x-7)5 2- ^ = (x-4)t(r5)^ - *• ^ = ^(1 + ^)^^^- 5. 2/ = x"n^. (Solve by two methods.) 6. 7/ = x^^°^. 7. s = (7< + 3)10^-2. 8. 2/ = xV'^. In Exercises 9-16 find the logarithmic derivative. 9. 2/ = e^'. 12. y = x". 10. y = x\ 13. y = ex". §84] EXPONENTIAL AND LOGARITHMIC FUNCTIONS 155 11. y = Zi' 14. 2/ = e*'+^ = ce**. 15. y = 10*'+^. 16. y = uv, where u and v are functions of x. 17. y = uvw, where u, v, and w are functions of x. Find t— Find y if its logarithmic derivative is: 22. A;. (1) (2) Equation (2) expresses the fact already noted in §82, as a characteristic property of the exponential function, viz., that the function increases at a rate proportional to itself. We can show, conversely, that if a function increases at a rate proportional to itself, it is an exponential function. Thus, let it be given that dy 18. 61 + 7. Ans. y = Ce ' X 19 J. X 20. - X 21. FkxY 84. Compound Interest Law. If y = Ce*S dy dt CA;e*' = ky. dt = ^y- (3> Then dy kdt logy = kt + C 2/ = e*'+c = e^e*'. Hence y = Cie*'. (4) When a function varies according to this law it is said to follow the "compound interest law." For, if a sum of money be placed at compound interest, its rate of increase, for any interest term, is proportional to the amount accumulated at the beginning of that term. The more frequently the interest is compounded the more nearly does the amount accumulated increase according to the exponential law. In many cases in nature the function decreases at a rate pro- 156 CALCULUS portional to itself. The compound interest law appears in this case in the form Ce~*', where A is a positive constant. For, if ^= -kv it follows that y = Ce-*«. Illustration 1. Newton's law of cooling states that the tem- perature of a heated body surrounded by a medium of constant temperature decreases at a rate proportional to the difference in temperature between the body and the medium. Let d denote the difference in temperature. Then ft = - ^^- (S) and e = Ce-*«. (6) The meaning of the constant C is seen at once on setting t = 0. It is the difference in temperature between the body and the medium at the time t = 0. If this initial difference in temperature is known, (6) gives the temperature of the body at any later instant. Call the initial difference in temperature ^o- (6) becomes e = doe-"'. (7) The time which is required for the difference in temperature to fall from di to 62 can be found from (7). Thus 01 = 0oe-*'i 62 = ^oe-*'» whence /.-<. = ^ log |. (8) This result could have been obtained directly from the differ- ential equation (5). Thus ..= -if. (9) §84] EXPONENTIAL AND LOGARITHMIC FUNCTIONS 157 Integrating the left-hand member between the limits ti and <2 and the right-hand member between the limits di and 62, f— iX e Illustration 2. Find the law of variation of the atmospheric pressure with height. Consider a column of air of unit cross section (Fig. 60). Denote height above sea level by h and the pressure on unit cross section at this height by p. The difference in pres- sure at C and D is the weight of the gas within the element of volume of height Ah. Thus Ap = — gp Ah, where p is the average density of the air in the volume CDEF. Then Fig. 60. and Ap Ah dp dh = - gp = - gp, (1) where p is the density at C. If the temperature is assumed con- stant, the air obeys Boyle's Law, pv = c, where v denotes the volume occupied bj'^ unit mass of air. Since mass _ 1^ _ P volume V c P = dh = - kp, 158 CALCULUS where * = ^ Integration gives [§84 logp = — kh + logCi, or When h = 0, p = Po, the pressure at sea level, and Ci = po- Hence p = poe~**. (2) If h is measured in meters and p in millimeters of mercury, k = risVir, (2) becomes p = 760e-iToT- (3) Exercises 1. A law for the velocity of chemical reactions states that the amount of chemical change per unit of time is proportional to the mass of changing substance present in the system. The rate at which the change takes place is proportional to the mass of the substance still unchanged. If q denotes the original mass, find an expression for the mass remaining unchanged after a time t has elapsed. 2. Assuming that the retardation of a boat moving in still water is proportional to the velocity, find the distance passed over in time t after the engine was shut off, if the boat was moving at the rate of 7 7 miles per hour at that time. Ans. s = t(1 — e *'). 3. The number of bacteria per cubic centimeter of culture increases under proper conditions at a rate proportional to the number present. Find an expression for the number present at the end of time t. Find the time required for the number per cubic centimeter to increase from 6i to 62- Does this time depend on the number present at the time t = 0? 4. A disk is rotating about a vertical axis in a liquid. If the retar- dation due to friction of the liquid is proportional to the angular velocity «, find w after t seconds if the initial angular velocity was wo. 6. If the disk of Exercise 4 is rotating very rapidly, the retardation is proportional to w^ Find w after t seconds if the initial angular velocity was wq. §85] EXPONENTIAL AND LOGARITHMIC FUNCTIONS 159 85. Relative Rate of Increase. If the rate of change of a func- tion is divided by the function itself, the quotient is the rate of change of the function per unit value of the function. This quotient has been called the relative rate of increase of the function. If a function varies according to the compound interest law, its relative rate of increase is constant, i.e., y dt One hundred times the relative rate of increase is the percent rate of increase. Thus if - t = 0.02, V dt ' the percent rate of increase is 2. This means that y increases 2 percent per unit time. Any of the Exercises 1-5 might have been stated in terms of the relative rate of increase of the function concerned. Exercises 1. Given that the intensity of light is diminished 2 percent by passing through one millimeter of glass, find the intensity / as a func- tion of t, the thickness of the glass through which the light passes. 2. The temperature of a body cooling according to Newton's Law fell from 30° to 18° in 6 minutes. Find the percent rate of decrease of temperature per minute. 86. Hyperbolic Functions. The engineering student is likely to meet in his reading certain functions called the hyperbolic functions. These functions present analogies with the circular functions and they are called hyperbolic sine, written sinh, hyper- bolic cosine, written cosh, and so on. These functions are defined by the equations: cosh X = — n — ' sech x = . , e' — e~' , smh X = — -X » csch x = 2 gx — g-x 6' — e~* tanh X = .. , ,_. ' coth x = odd functions e* 4- e~* tanh x cosh X and sech x are even functions, while the remaining four are 160 CALCULUS [§87 Exercises 1. By making use of the definitions the student will show that the following identities hold. They are analogous to those satisfied by the circular functions. cosh^x — sinh^x = 1. 1 — tanh*x = sech^x. 2. Show by the use of the defining equations that : d cosh X dx d sinh X dx d tanh x dx d coth X dx d sech X dx d csch X sinh X, = cosh X. = sech^ X. = — csch^ X. = — sech X tanh x. = — csch X coth X. dx 3. Sketch the curves y = cosh x, y = sinh x, and y = tanh x. 87. Inverse Hyperbolic Functions. The logarithms of certain functions can be expressed in terms of inverse hyperbolic functions Let y = sinh~* x. e« — e~y J/ — Buiii (/ — 2 or e'^y — 2xey — 1=0, whence c = X ± Va;^ + 1. The minus sign cannot be taken since e" is always positive. Hence e" = a; + Vx^ + 1 , and y = sinh-' x = log (x + Va;^ + !)• 1. Show that Since §88] EXPONENTIAL AND LOGARITHMIC FUNCTIONS 161 Exercises cosh~i a; = log (x + va;" — 1). X - -y/x^ - 1 = 7 > X + Vx'^ - 1 log (x - a/x" - 1) = - log {x + \/x2 - 1). Therefore cosh~ia; = + log (x + 's/x^ — 1). The inverse hyperbolic cosine is then not single valued. Two values of cosh~i x, equal numerically but of opposite sign, correspond to each value of x greater than 1. 2. Show that: 1 + x tanh'^x = § log , _ > if x'^ < 1; X + 1 coth~i X = 2 log _ ^ » if x^ > 1 ; 1 -f. a/1 _ a;2 sech~* X = ± log > if < x ^ 1 ; and , _. , 1 + Vx^ + 1 .. csch ^x = log > if X > 0: ^ X ' csch ^ X = log > if X < 0. The student is not advised to memorize the formulas of this and the preceding sections at this stage in his course, but to acquire sufficient familiarity with the hyperbolic functions to enable him to operate with these functions by referring to the definitions and formulas given here and to others that he will find in mathematical tables. 88. The Catenary. Let AOB, Fig. 61 a, be a cable suspended from the points A and B and carrying only its own weight. Let us find the equation of the curve assumed by the cable, consider- ing it homogeneous. We shall assume that the curve has a vertical line of symmetry, OY, and that the tangent line drawn to the curve at is horizontal. Take 07 as the F-axis. Imagine a portion of the curve, OP, of length s, cut free. To hold this portion in equilibrium the forces // and T, Fig. 61 h, must be introduced at the cut ends. n 162 CALCULUS H and T are, respectively, equal to the tension in the cable at the points and P and they act in the direction of the tangent lines drawn to the curve at these points. The portion of the cable OP, Fig. 61 b, is in equilibrium. Hence H', the horizontal com- ponent of T, is equal to H. V, the vertical component of T, must balance the weight of the portion OP of the cable. Hence V = sw, where w is the weight of a unit length of the cable. From Fig. 61 6, it is seen that dy _ V^ _ V^ _ ws dx~ H' ~ H ~ H' Let w _ 1 H ~ a Then dy s di " a' (^) This differential equation involves three variables, viz., x, y, and 8. s may be eliminated by differentiating and substituting for ^ its value, Thus dx a dx a \^ + W EXPONENTIAL AND LOGARITHMIC FUNCTIONS 163 The equation now involves only two variables and may be written , dy ok upon -j- tion (2) is = -dx. (2) dv If we look upon -r- as the variable u, the left-hand side of equa- du vTTw2 whose integral is log (u + \/l -|- u^). (See Illustration 18, §82.) Integrating (2), When a; = 0, ^ = 0. Hence C = and (3) becomes dv From the symmetry of the curve -r- changes sign when x is re- placed by —X. Then from (4), Subtracting (5) from (4), 2g=e«-e"a, (6) or ^ = sinh? (7) dx a Integrating (7), y = a cosh - -f- d- (8) If the origin is taken a units below the point 0, Fig. 61 a, y = a when x = 0, and C2 = 0. Hence y = a cosh — (9) 164 CALCULUS This is the equation of the curve assumed by the cable. It is called the catenary. Equation (9) can be written Y = cosh X, (10) where r = ^ and X = ?• a a The constant a depends upon the tautness of the cable. Equa- tion (10) shows that the curve y = cosh z if magnified the proper number of diameters will fit any cable hanging under its own weight. The length of OP can be found by substituting in formula 2, §63, du the value of j- given by (7), and integrating. -^' ds = « / 1 + sinh' - dx a 1C = cosh- dx. a X s = a sinh + C3. a Since s is measured from the point where the curve crosses the y-axis, s = when x = 0. Hence Cj = and s= a sinh -• (11) a Exercises 1. If the two supports A and B, Fig. 61, are on a level, L feet apart, and if the sag is d feet, show that the tension, T, in the cable at the points A and B is T = w(a + d). 2. Beginning with equation (6) find expressions for y and s without making use of hyperbolic functions. 3. If the cable. Fig. 61, is drawn very taut, show that the equation of its curve is approximately X2 y = 2a if the origin of coordinates is taken at the lowest point of the cable. Hint. Begin with equation (2) and note that i-r-j is small com- pared with 1. CHAPTER X MAXIMA AND MINIMA In previous chapters maximum and minimum values of func- tions have been found by making use of the derivative. Besides this method several others which do not involve the use of the derivative may, at times, be used to advantage. 89. The Maximum or Minimum of y = ax^ + j8x + 7. In elementary analysis the student learned that y = az^ + j8a; + 7 represents a parabola with its axis parallel to the F-axis, and that the equation can be put in the form y = a(x — p)^ -\- q. The point (p, q) is the vertex of the parabola. If a is positive, the vertex is a minimum point, if negative, a maximum point of the curve. Let y = Sx^ - 12x + 19. y = 3ix- 2)2 H- 7. The last equation shows at once that the minimum value of the function is 7 and that it occurs when x = 2. Exercises Find the maximum or minimum values of the following: 1. y = 3x2 _ 2x + 1. 2. y = 3x - 2x2 + 1. 3. y = 3x2 _^ 7a., 4. If a body is thrown vertically upward with an initial velocity of a feet per second, its height h in feet at the end of t seconds is given by h = at - 16.1<2. To what height will the body rise if thrown with an initial velocity of 32.2 feet per second? When will it reach this height? 90. The Function a cos x + b sin x. The function a cos x + 6 sin X is of frequent occurrence. If it is put in the form of 165 166 CALCULUS ^90 the product of a constant by the cosine of a variable angle, the maximum and minimum values can be found at once. Thus cos X + —r= a cos X -\-hsmx = ^aP' + h''' a b Now, — / and — , 'Va^ + b^ Va^ + b^ sine, respectively, of an angle a. For if P, Fig. 62, be the point (a, 6) and the angle POX be a, Va^ + 62 sin X [Va^ + b^ may be regarded as the cosine and and Va^ + b^* b Va^ + b^ Fig. 62. Hence a cos X -\- b sin x = y/a^ -\- b^ (cos x cos a + sin x sin a), or a cos X -r b sinx = -x/a^ -j- b^ cos (x — a). (1) The quadrant in which ce lies will be determined by the signs of a and b. a is in the first quadrant if a is positive and 6 is positive. a is in the second quadrant if a is negative and b is positive. a is in the third quadrant if a is negative and b is negative. a is in the fourth quadrant if a is positive and b is negative. In polar coordinates equation (1) shows that the function a cos X + bsinx is represented by a circle passing through the pole, of diameter -y/a^ + b^, and with its center on the line making an angle a with the polar axis. The right-hand side of equation (1) shows that the function is represented graphically in rectangular coordinates by a cosine curve of amplitude -y/a^ -f b^. Thus, the maximum value of a cos X + & sin a; is \/a^ + b^ and occurs when x = a. The mini- mum value of the function is — \/a'^ + b^ and occurs when 2 = a -|- X. Two examples giving rise to this function are solved below. §90] MAXIMA AND MINIMA 167 Illustration 1. The weight W, Fig. 63, rests upon the horizontal surface AB. P is the force, inclined at an angle 6 with the horizontal, which will just cause the weight to slide over the plane. The problem is to find the angle 6 for which P will be a minimum. The coefficient of friction is denoted by ^i. The normal pressure, N, between the weight and the plane is (]F — P sin 0), the difference between W and the vertical com- ponent of P. The force of friction, F, is then fi {W — P sind). The horizontal component of P equals F. Hence niW - Psind) = Pcosd, or ^ = cos0+Atsine. (2) uW Since ix and W are constants,-p- is a maximum when and only when P is a minimum. Hence to find the minimum value of P uW we may find the maximum value of -p- and multiply its reciprocal hy fiW W p B A < — i:i^ i tiw^m i B N Fig. 63. Fig. 64. From (2), ^ = Vl + M^' cos (0 - a), where a is an angle in the first quadrant whose tangent is /x, the coefficient of friction. Therefore, when 6 is acute and equal to tan~^ /x, JP is a minimum and equal to , Vl + /X'' Illustration 2. A weight, W, Fig. 64, rests upon the inclined plane AB. Find d so that P, the force which will just cause W to move up the plane, shall be a minimum. The normal pressure, N, between the weight and the plane AB 168 CALCULUS I §91 is equal to Tf cos (3 - P sin 6. Then F, the force of friction, is equal to H iW cos iS - P sin 6), where /t is the coefficient of friction. We have then F = m(^ cos |8 -Psin0). Since the force of friction must balance the components of P and W parallel to the plane AB, we have H{W cos i3 - P sin d) = (P cos e -W sin /3). Hence Win cos j8 + sin /3) — ^^- p ^-^ = cos + /i sm 6, or Tr(^cos^ + sin/?) ^^^^-^^^^^^_^^ (3^ where a is the acute angle whose tangent is n. Thus, the left- hand side of (3) is a maximum, and P a minimum, when d is acute and equal to tan~^ ju. „, . . . , f p • .K Tr(MCos^ + sini8) The mmimum value of P is then . Exercise In Fig. 64, find the minimum force, P, and the angle between its line of action and AB, which will just prevent the weight W from slid- ing down the plane. 91. The Function mx + -y/a^ — x^. Frequently problems in maxima and minima lead to functions of the form mx ± ■s/a'^ — x^. The curve for y = mx ± y/a^ — x"^ (1) can be obtained by shearing the circle y = + \/a^ — x^ in the line y = mx. Every ordinate of the circle to the right of the F-axis is increased (or decreased if m is negative) by an amount propor- tional to the distance from the F-axis. To the left of the F-axis the ordinates are decreased if m is positive and increased if m is negative. The maximum value of y is easily found by placing X = a cos t. (2) §93] MAXIMA AND MINIMA 169 Then from (1) y = a(m cos t + sin t) (3) = aVl + m^ cos (t — a), (4) where a = tan"^ — The maximum value of y occurs when X = a cos a = vTT 92. Maxima and Minima by Limits of Curve. In case f(x, y) = is of the second degree in x and y, and in a few other cases, the maximum and minimum values of y can be found by determining when X changes from real to complex values. The method will be illustrated by an example. Let ^ g^ + e ^ ~ 2x + l" Then x = y ± V{y + S){y-2). (1) From equation (1) it is seen that for values of y greater than 2 or less than — 3, x has two distinct real values. When y = 2 or — 3, X has two equal real values. When — 3 < ?/ < 2, a; is imagi- nary. This shows that the line y = c meets the curve in two distinct points if it is more than two units above or more than three units below the X-axis; that it is tangent to the curve when two units above and when three units below the X-axis, and that it does not cut the curve when it falls within the limits two units above and three units below the X-axis. Hence the func- tion has a minimum value 2 and a maximum value —3. Exercises 1. Find the luaximum and minimum values of x" - 2x + 19 2z +5 2. Find the maximum rectangle which can be inscribed in a circle of radius 10. 93. Maxima and Minima Determined by the Derivative. The first derivative has been used to determine the value of the argu- 170 CALCULUS [§93 ment corresponding to maxima and minima of functions. Im- mediately to the left of a maximum point the function is increas- ing with X and consequently the first derivative is positive. On the other hand, immediately to the right of such a point the func- tion is decreasing as x increases and the first derivative is negative. Similarly it follows that the first derivative is negative im- mediately to the left and positive immediately to the right of a minimum point. In both cases the first derivative changes sign as the independent variable passes through the value for which the function has a maximum or a minimum value. This change of sign may take place in a number of ways. Illustration 1. Thus, in the case of the function y = x^ -2x + 7, the derivative, I = 2. - 2 . 2(. - 1), is negative to the left and positive to the right of the line x = 1, dy When a; = 1, ^ = and the curve has a horizontal tangent. In the vicinity of this point the curve has the shape shown in Fig. 65. At first thought it might appear that if the first derivative is negative to the left and positive to the right of a certain point, it certainly must become zero at this point. This is, however, by no means the case, as the next illustration will show. Illustration 2. y = 4 + (x — 1)^. Although the minimum value of this function can be determined at once by noting that it represents a semi-cubical parabola with its vertex at (1, 4), the problem will be worked by the method of the calculus for illustra- tive purposes. The derivative, dy^ 2 ^ ^^ 3(x-l)*' is negative when x < 1 and positive when x > 1. Hence the function is decreasing to the left and increasing to the right of §93] MAXIMA AND MINIMA 171 a; = 1. When x = 1, ?/ = 4. This value is a minimum value of the function. For z = I the derivative does not exist, as the denominator becomes zero. Let us see what really happens in the vicinity of x = 1. As x approaches 1 from the left, j- takes on larger and larger negative values. The form of the curve to the left and in the immediate vicinity of the point (1, 4) is some- thing like that shown in Fig. 66. The line x = 1 is tangent to the curve at this point. As X approaches 1 from the right, i.e., through decreasing values, the derivative becomes larger and larger. The form of the curve to the right of the line x = 1 is also shown in Fig. 66. The line X = 1 is also tangent to the portion of the curve obtained by allow- ing X to approach 1 from the right. Fig. C6. Fig. 67. It is now apparent that the first derivative may change sign without passing through zero. In the above illustration it changes sign by becoming infinite. The first derivative may change sign in still another way as illustrated by the curve of Fig. 67. Let us suppose that the de- rivative approaches — 1 as x approaches a from the left, and the value +1 as X approaches a from the right. The function has a minimum value at the point P, for the derivative changes from minus to plus as x increases through the value a and consequently the function is decreasing to the left and increasing to the right of X = a. The essential thing at a minimum point is that the derivative changes sign from minus to plus, and at a maximum point that it changes sign from plus to minus. 172 CALCULUS [§95 A derivative which is continuous at a maximum or a minimum point changes sign by passing through zero. But it may change sign by becoming infinite, as the second illustration shows, or by becoming otherwise discontinuous as explained above. This last type is of rare occurrence and will not be referred to again. Illiistration 3. y = x^ -^ 3. The derivative of this function, -p = 3x^, is positive for all values of x except x = 0, when it is zero. The function is increasing for all these values of x. At this point, (0, 3), there is a horizontal tangent but the function has neither a maximum nor a minimum at the point, for it in- creases up to the value 3 for x = and then continues to increase to the right. This illustration brings out clearly the fact that there is no reason for assuming that a function has a maximum or a minimum value at a point where the first derivative is zero. What kind of a point is the point (0, 3) ? 94. Second-Derivative Test for Maxima and Minima. In the first of the three types of maximum or minimum points considered in §93, the first derivative changes continuously from positive to negative values or vice versa. For a maximum point of this type the curve is concave downward and the second derivative is negative at such a point. For a minimum point the curve is concave upward and the second derivative is positive. A con- venient test for the behavior of a function at a point where the first derivative is zero is then, to substitute the abscissa of this point in the expression for the second derivative. If the second derivative is positive the point is a minimum point; if negative, a maximum point. If the second derivative is zero, the test fails. This test also fails for maximum or minimum points where the first derivative is discontinuous. Examine the curves y = x», y = X*, 95. Study of a Function by Means of its Derivatives. The following is a summary of the application of the first and second derivatives to tracing the curve representing a function: §95] MAXIMA AND MINIMA 173 1. The function is increasing if the first derivative is positive, and decreasing if it is negative. 2. To find maximum and minimum points find the values of X for which the first derivative becomes zero or infinite. If the derivative changes sign at any of these points, the correspond- ing point is a maximum or minimum point according as the change is from plus to minus or vice versa. Points at which the first derivative is equal to zero can also bo tested by substituting the abscissa of the points in the second derivative. If the second derivative is positive, the point is a minimum point, if negative, a maximum point. 3. Points of inflection are found by determining where the second derivative changes sign. As in the case of the first derivative, the change in sign can take place through zero or infinity. If the change is from positive to negative values the curve changes from being concave upward to being concave downward. The abscissas of the points at which the first and the second derivatives become zero or infinite we shall call the critical values. These values and these alone need be tested in study- ing the behavior of an ordinary curve. The investigation of a curve by means of its derivatives can be put in the tabulated form shown in the following illustrative examples: 1. y (See Fig. 33.) dy dx d^ dx^ — X. dy dx = when x = 0. -1-^ = when x X dhj dx^ dy dx y X <0 X > <0 >0 >0 >o Concave downward, increasing Concave upward, increasing 174 CALCULUS ^96 Here (0, 0) is a point of inflection. There is neither a maxi- mum nor a minimum point. 2. 1/ = ix3 - x^ + Ix + 2. (See Fig. 34.) g = ix^ - 2x + I = X i(:r - Dix - 3). 2. dv -^ = when x ax = when x 1,3. 2. X dy dx y X < 2 X > 2 <o > Decreasing Increasing Concave downward Concave upward X < 1 1 <x < 3 X > 3 > < > Increasing Decreasing Increasing 1 2 3 8 3 i 2 2 (2, I) Point of inflection. (1, f) Maximum point. (3, 2) Minimum point. Apply second derivative test for x = 1 and x = 3. 96. Applications of Maxima and Minima. In solving problems involving maxima and minima the first step is to set up from the conditions of the problem the function whose maximum or mini- mum value is sought. Frequently the function will be expressed, at first, in terms of two or more variables. Usually, however, there is a relation between these variables, and the function can be §96] MAXIMA AND MINIMA 175 expressed in terms of a single variable. After this has been done the maximum or minimum values can be found. Exercises 1. Equal squares are cut from the comers of a rectangular piece of tin 30 by 20 inches. The rectangular projections are then turned up forming the sides of an open box. Find the size of the squares cut out if the volume of the box is a maximum. 2. A man who is in a boat 3 miles from the nearest point, A, of a straight shore wishes to reach, in the shortest possible time, a point B on the shore which is 6 miles from A. Find the point of the shore toward which he should row, if he can row at the rate of 3 miles per hour and walk at the rate of 5 miles per hour. 3. The horizontal component of the tension in the guy wire BC, Fig. 68, is to balance the horizontal pull P. If the strength of the wire varies as its cross section, and if its cost varies as its weight, find the angle such that the cost of the guy wire shall be a minimum. B Fig. 68. Fig. 69. 4. Find the length of the shortest beam that can be used to brace a wall if the beam passes over a second wall 6 feet high and 8 feet from the first. 5. A steel girder 30 feet long is moved on rollers along a passageway 10 feet wide, and through the door AB, Fig. 69, at the end of the pas- sageway. Neglecting the width of the girder, how wide must the door be in order to allow the girder to pass through? 6. A sign 10 feet high is fastened to the side of a building so that the lower edge is 25 feet from the ground. How far from the building should an observer on the ground stand in order that he may see the sign to the best advantage, i.e., in order that the angle at his eye sub- tended by the sign may be the greatest possible? The observer's eye is 5 1 feet from the ground. 7. A man in a launch is m miles from the nearest point A of a straight shore. Toward what point on the shore should he head his boat in order to reach, in the shortest possible time, an inland point 176 CALCULUS [§96 whose distance from the nearest point B of the shore is n miles? The man can run the boat Vi miles per hour and can walk Vj miles per hour. The distance AB is p miles. Ans. Toward a point such that sin 61 sin dt Vl V2 where 0i and 62 are the angles made by the paths of the man with the normal to the beach. It will be noticed that the path taken by the man is similar to that followed by a ray of light in passing from one medium to another with a different index of refraction. -.25 .5 1 1.5 Fig. 70. 8. A man in a launch is m miles from the nearest point A of a straight shore. He wishes to touch shore and reach, in the shortest possible time, a second point on the lake whose distance from the nearest point B on shore is n miles. In what direction must he head his boat if the distance AB is p miles? The path taken by the man is similar to the path of a ray of light reflected by a plane surface. 9. It is desired to make a gutter, whose cross section shall be a MAXIMA AND MINIMA 177 segment of a circle, by bending a strip of tin of width a. Find the radius of the cross section of maximum carrying capacity. 10. A sector is cut from a circular piece of tin. The cut edges of the remaining portion of the sheet are then brought together to form a cone. Find the angle of the sector to be cut out in order that the volume of the cone shall be a maximum. 11. The stiffness of a rectangular beam varies as its breadth and as the cube of its depth. Find the dimensions of the stiffest beam which can be cut from a circular log 12 inches in diameter. 12. The strength of a rectangular beam varies as its breadth and as the square of its depth. Find the dimensions of the strongest beam which can be cut from a circular log 12 inches in diameter. 1.5 L26 .75 (4) .25 k^^ 3^(2) (1) 2/ = e"* +0.56-^"* (2) y=e- (3) 2/ = e ^-o.ie (6) 2/ = e-* -1.56 ^ >A(3) p^ \, |(6) \ s^ — - 1 1.5 Fig. 71. 2.6 13. Consider the sum y = ox" + bx*" for positive values of x only. First, if n and r are of like sign, show that: (1) a maximum or a minimum value exists if a and b are of unlike sign; (2) neither a maximum nor a minimum value exists if a and b are of like sign. Second, discuss the same cases if n and r have opposite signs. 14. Determine the exact values of the maxima shown in Figs. 70 and 71. Hint. Consider first the general case 12 y e~' — ae CHAPTER XI POLAR COORDINATES 97. Direction of Curve in Polar Coordinates. Let BPQ, Fig. 72, be a curve referred to as pole and OA as polar axis. Let P be any point of the curve and let PT be a tangent to the curve at this point. Let PS be the radius vector of the point P, produced. A point describing the curve, when at-P, moves in the direction PT. This direction is given by the angle ^ through which the radius vector produced must rotate in a posi- tive direction about P, in order to be- come coincident with the tangent line. An expression for tan^ will now be found. Let Q, Fig. 72, be a second point of the curve. PR is perpendicu- lar to OQ, and PM is a circular arc with as center and radius OP = p. tan^= iJ^otani2QP= iJS'o ^. (D The infinitesimals PR and RQ can be replaced by PM and MQ, respectively, if (see §60) lim PR A0=O PM Fig. 72. 1 (2) and = 1. lim RQ Equation (2) is true by equation (3), §56 (3) follows: lim RQ^ Afl-.0 MQ (3) The proof of equation lim RM + MQ Ae=o MQ Urn p(l -cos Ag) + A p A9=0 ^p , , lim p(l-cosA0) Ae 178 Ad Ap §97] Hence since POLAR COORDINATES 179 lim RQ = 1. Ae=o MQ lim 1 — cos Ad A9=0 ^0 From (1), (2), and (3) it follows that = 0. , , lim PM Hence lim pAg tan^ dp tan^ = ^- dd (4) This formula* can be easily remembered if the sides of the tri- angular figure MQP, Fig. 72, are thought of as straight lines, and d sin ' This formula enables us to give another proof for — -yz — . In polar codrdinates p = sin 8 represents a circle, Fig. 73. By geometry, x// = e. Then p sin tan^ = tanfl=^ - ^. de dp de "■ d sin de de cos 8, 180 CALCULUS [§97 the angle MQP as equal to \p. Then the tangent of rp would be de p_ dp dd dp dp dv Formula (4) corresponds io -r = tan t in rectangular coordi- nates. Illustration L If p = e"*, dd = «^ and tan \I/ = -, a, constant. ^ a' Illustration 2. Find the equation of the family of curves for which the angle between the radius vector produced and the tangent line is a constant. tan yp = k. or Integrating or dp dd k, dp ; dd _ P 1 k dp P >• IP = 1-- P = «:— = e e*, P = Kek, where K is an arbitrary constant. ^98] POLAR COORDINATES 181 Exercises Find tan \p for each of the following curves : 1 '^. 4. p = a(l — cos e). 5 o 2. p = ad. '' 1 - cos » 3. p = e"'. 6. p = o cos (0 — a). 98. Differential of Arc: Polar Coordinates. We. shall now ds find an expression for -v^ in polar coordinates. From Fig. 72, dd (chord PQ) 2 = (PRy + (RQ^. From which lira /chord PQV lira /PRy , lira /RQV Replacing chord PQ by arc PQ = As, PR by PM = pA0, and RQ by MQ = Ap, (1) (2) (3) It corresponds to (ds)^ = (da-)^ + (%)'' in rectangular coordinates. It can be remembered easily by the help of the triangle MQP, Fig. 72. The length of the curve can be expressed as a definite integral. Thus: (See Fig. 74) lim -^ _,^ s = A« = 2Lf P^ lira / A9 = 1 Y lim fpAey , 1 Therefore [de) = ^ + [de) and dS = Jp2+(^P)W This formula can be written (ds)2 = p2(d0)2 + (dp)^ 182 CALCULUS §98] = a/™o % ViPRV + {RQy B = /3 -J'i-0 D'VQV (ff)'- = a = fi Illustration. Find the entire length of the curve p = a(l — cos 0). This curve is symmetrical with respect to the polar axis. The length of the upper half will be found and multiplied by 2. dp = j Va\i - cos ey + a2 sin2 d dd =^^de a sin -H" rf0 = — 4a cos — = 4a. ^ lo s = 8a. Exercises 1. Find the entire length of the curve p = 2a sin 9. 2. Find the entire length of the curve p = a(l — sin 6). a 3. Find the entire length of the curve p = a sin' -^• o 4. Find the length of p = e°* between the points corresponding to 6=0 and = v. Also between the points corresponding to = and e =Z- §99] POLAR COORDINATES 183 5. Prove formula (3) directly from and X = p cos d, y = p sin e, ds = Vidxy + (dj/)2. 99. Area : Polar Coordinates. Find the area bounded by the curve p = fid) and the radii vectores 6 = a and 6 = ^. We seek the area BOC, Fig. 74. Draw radii vectores dividing the angle BOC into n equal parts Ad. Let POQ be a tj'pical one of the n portions into which the area is divided by these radii. The angle POQ is Ad. The line OP makes an angle 6 with the initial line OA, and ^^°- '^' its length is p = f(d). Denote the area of BOC by A. lim o7 ^ =nT^XPOQ^ (1) Replace^ POQ by the circular sector POR whose area is ^p^'A^. Then ^ s or 'I P^dd, A = h\ [mv dd. Exercises (3) 1. Find the area bounded by the curve p = 2o sin e. 2. Find the area bounded by the cardioid p = 2a(l — cos 6). , 1 + cos 29 Hint. cos*0 = ^ iLetA^ = OPQ (Fig. 74). PR and QS are arcs of circles. Then OPR < A^ < OSQ, Jp«Ae <AA< i(p + Ap)'AJ. 184 CALCULUS [§99 3. Find the area bounded by p = 2a(l + sin e). 4. Find the area bounded by one loop of p = 10 cos 2d. 5. Find the area bounded by one loop of p = 10 sin 26, 6. Find the area bounded by one loop of p = a cos Sd. 7. Find the area bounded by one loop of p^ = 10 cos 29. 8. Find the area swept out by the radius vector of the curve p = ad, as varies from to 2ir. 9. Find the area bounded by the radii vectores = 2, ^ = ir, and the curve p = -r^- 10. Find the area bounded by the radii vectores = 0, 6 = ^ and the curve p = 5^^. CHAPTER XII INTEGRATION 100. Formulas. In Chapters III, VI and VII the following formulas of integration, with the exception of (19), have been used. They are collected here foj reference, and should be memorized by the student. 1. fw du = — 7-r w»+i + C, if w 7^ - 1. J n + 1 3. J e» rfw = e» + C. 4. I a" du = , a» + C. J log, a 5. j sin udu = — cos u -\- C. 6. J cos w dw = sin w + C. 7. I sec'^ udu = tan u + C. 8. I csc'^ udu = — cot w + C. 9. I sec u tan udu = sec m + C 10. I esc M cot udu = — esc w + C 11. J tan udu = log sec w + C. 12. J cot udu = log sin w + C 13. J sec udu = log (sec w + tan u) + C 14. { CSC udu = — log (esc w + cot u) + C. 15. I —r = sm-i - + C. 185 186 CALCULUS 16. du a^ + u^ = - tan-i - + C. a a [§100 17. I .- = - sec-i - + C. J u\/u'^ — a2 « « 18. I ,J^ = log (m + Vu^ ± a^) + C. r du ^ 1^ J u^-a^~ 2a 19. Iri^.-^loK^ + C.ifX- = K- log — j \- C,itu < a. 2a ^ a + u Formula (19) is proved as follows: .^^lU Ll u2 — qZ 2a \u — a u-\- aj f du ^ i_ rr_i 1 1 J u^ - a^ 2aJ [u-a w + a J " _ 1 r du 1 r du ~ 2a J u — a 2a J u-\- a = 2" log {u - a) - 2^ log (w + a) -f C. 1 , u-a ^ = ^ log — , h C. 2a ^ u + a This formula leads to the logarithm of a negative number if m < a. To obtain a formula for this case write ^_ = ir__i Li. u^ — a^ 2a L a — u a-\-uJ Then / du 1 , a — M . _ = o:; log :,-n-7: + C- ■u^ — a^ 2a ^ a + w Exercises r xdx r * J Vie - z2* ^* J V^ ^* J Vie - X*" *' J ^' + 16 2-16 dx §100] INTEGRATION 187 r dx • J16-X I 16 xdx x^ - 16* dx xy/x^ - 16 9. fcot 1i dt. r {x+ a)dx ^ J ^' + 2ax 11. fix^ - 16)^ X dx. 12. fsin (2x - 3) dx. 13. fsec^ (5a + 2) da. 25. I e2 +e~2 dx. 26. la^'' dx. 27. r(2x + 4)''-" dx ^^' J 7x* + ll 30. I 4. Jsec (2(? + 4) tan (20 + 4) de. 5. fcsc^ (3 - 2</.) d<^. 7. fe^^'^^ainede. C 31^ dt J 30<'+ 13' 9. JiVc - Vxy dx. 20. J \/3 + 4x dx. /I e^dy. 23. fe*'*" (2^ +2) sec2(2x + 3) dx. dx. Divide numerator by denominator. 31. le^'dx. dx / dx 4x2 _ / J Vie - 9x2" 35. ftan (3a + 4) da. 33 34 9 cos X dx 4 + 3 sin X dx cos" (3x - 2) 36. Htan e + cot ey dd = tan e - cotB + C. 39. Jcos (3< - 4) dt. „ 40 r ^*- 38. J cot (5< - 8) dt. J ^ - *2/' /■ 37. I (sin -z — cos 50) dd. 188 CALCULUS [§100 dz 61. fcos^ (3x - 2) sin (3x - 2) dx. 62. I sec* (9 - 7x) dx. 63. I e 3 sec^^dx. •{. 63. 43. J tan (2x - 5) dx. J 44. Jsec (2y + 4) dt/. 64. J tan* x sec* a: dx. 46. fcsc (2t/ — 7) dy. 65. J cos* 3x sin 3a; dz. 46. J cot (3< + 11) d/. 66. ftan^ 5x sec* 5xdx. 47. I sec* (^ — 5]dx. 67. I sec* x tan xdx. 48. jTcos (3 - 2x) dx. «8. J csc^ x cot x dx. Aa r 2x + 5 ^ 69. fx tan (2x* - 5) dx. **• I x* + 5x + 4r^- % / _ I sin X cos X dx 62 / <^^ 72. (e* —^dx. V9<* - 4 J * / <^^ 73. fiJ - x»)» dx. V4-9/*- J <\/9<* - 4 "^ / tdt 75. ft^Qt^ - 17 dt. V9i^^4: r ^dx r tdt ^^J j9t*-4 n . I sec 5x dx. ' j i V9 -X* sin 5x dx 57. I sec 5x dx. ' / ^ ^*^^ ^^ "I" ^^ 68. fsin (cot + a) dt. 78. fe^''' ^^ cos 3x dx. 69. Jcos* 4x sin 4x dx. 79. fe^' + <^ + '^{x + 3) dx. 60. j sin* (x +3) cos (x + 3) dx. 80. | cos 5x sin 3x dx. INTEGRATION 83. Jsin 3t cos 4t dt. 84. I sin«5«d<. §100] INTEGRATION 189 81. I cos 3x cos 5x dx. 82. I sin 7x sin 4x dx. r 85. I sin mt cos n< di, where m and n are integers. What is the value if m = n? 86. I cos (3w< + a) sin (3w< + a)dt. 'I? 88 89 90 2x + 3 , x^ + 3x^+7 x* + 9 •J 3^^^^- ■ /3x-f-2 , 91. fsin* 5fl cos 56 dO. 92. I sin^ X dx. 93. J xVie - x2 dx. 94. I sec2 (l + 2J dx. 95.jy^^dt. 96. I sec 2 4x tan 4x dx. 97. I \ dx. 109. fsec {3<l> - 2) tan (30 - 2) d,/.. 110. rtanH2x - 1) sec* C2x - 1) dx. dx V5 - 7x2 dx VSx^ - 5 00. I sin 4x cos 6x dx. 01. f(V^-V^)'dx. '2x + 3 2x + 7 dx dx. V3x +2 04. I sin* 2x cos 2x dx. 05. I Vsinx cos x dx. 06. fe-^dt. 07. r(2x - 5)^dx. 08. / sec odx. 190 CALCULUS [§101 "^- j l + tan3x ^^- "8-j3^M^4 112. fv2^^3idx. 119. [ —y^ •^ J yVsy dy 113. ftan (5 - 2x) dx. f* vdv J 120 I —^=^ 114 116 f .i. J Vs.' -7 J 5 - 3x2 ]L21. I sec^ tan e dd. • J ¥~^7 "2- J VlP^' 116. I^^dx. 123. |sm'|cos|dx. /x + 4 C X — ; ?; <ix. 124. I cos 2xsin o<ia;. x^ - 9 J 2 101. Integration of Expressions Containing ax^ + bx + c, by Completing the Square. Illustration 1. / dx ^ r dx 1_ J X + 2 X2 + 4X + 9 J(x + 2)2+5 VS VS Illustration 2. /rfx r dx C dx V3 + 4x -"x^ ~ J VS - (x2 - 4x) ~ J \/3 + 4-(x-2)2 dx . X — 2 ■J: \/7 - (x - 2)2 \/7 Illustration 3. + C. r dx _ 1 r J7x2 + 3x + ll IJ x^ + ^x + tEit + V - ~lU dx 1 14 ^ _, x + -A- , ^ s — ;== tan * .> . + C ' V^99 V^99^ 14 2 tan -. lii+3 + c. 299 \/299 §101] INTEGRATION 191 Illustration 4. dx r dx i_ r dx 1^ r J VQ+2x-3x^~ V3J V2+lx-x^~VSJ V2-{x^-l^ 1 C dx 1 . _. x-l = — 7= I — ■— = — -7= sin ■* — 7= + C \/3j VY -ix-\r \/3 VY — ;- Sin^ ;p:r— + C. V3 \/l9 Exercises J x^ + 6x +25 J ^- J x2 - 6x + 5 J 5x^ - 4. f-^^ 9. f J A/2a;2 + 2x - 3 J V2 + 3t - 2t2 dx 8x + l dt Vl +2t + 2t2 2x2 _|. 5x - 3 ,„ r 2x-5 ^ f 2x+6-ll _, 1°- Jx^ + 6x+25^^ =J^2+6x+25'^^ / (2x + 6)dx _ r___^5___ x2 + 6x + 25 J x2 + 6x + 25* J V2x2^+ 2x - 3 J VS - 4x - x2 /4x + 1 1 C ^^-p2^-dx. 14. J- 16 r ^^ 18. f ^^ ' J xV3l^ 6x + 5x2 J V4J/2 + 122/ - 7 17. f-^-=' 19. fg^- dx 1 15. I ^ Substitute x = -• xv/2x2 + 3x - 2 ^ (Zx 24x -9 192 CALCULUS [§102 20. I 23. I . ^ =' J V8 + 12x - 4x» J x\/^+12x - 7x2 J \/8 + 12x - 4x2 J a X VSx^ + 12x - 4 / 22. 16x2 - 24x + 24 102. Integrals Containing Fractional Powers of x or of a + bx. Illustration 1. /: x^ — x^ — ax. 3 +4 Let X = 2*. Then dx = 62^ dz, and J ^i + 4 ^ j0^ + 4' ^" J^^ + 4 '''• The integration can be performed after dividing the numerator by the denominator until the degree of the remainder is less than 2. After integration replace z by x^. Illustration 2. \x + 2)^ + 4 /! dx. (x + 2)^-3 Let X + 2 = z<. Then r(x + 2)U4 f(?!+-41^'rf. Divide the numerator by the denominator. The integration can readily be performed. After integration replace 2 by (x + 2)*. In general if fractional powers of a single linear expressior, a + bx, occur under the integral sign, let a -{- bx = z", where n is the least common denominator of the exponents of a + bx. The linear expression a + bx reduces to x when a = and 6=1. See, for example, Illustration 1. Exercises §103] INTEGRATION 193 f. ■■/' ^„ , 3x- 2 , 10. I . dx. x\/2x + 3 (a;i + a;3) dx. 12. I . dx. Sx\/2x - 3 14. J 1 + (X + 2)^ 13. I v-^-T--^- -rx ^^^ 1 + (X + 2) Vx — 3 dx P x + 4 x(l + x^) 15. I ~" ' T dx. jg rV2x + 3 dx. 3x - 2 103. Integrals of Powers of Trigonometric Functions. (a) I sin" X cos" x dx where at least one of the exponents is an odd positive integer. This includes I sin"» x dx and J cos" x dx where the exponents are odd. Illustration 1. J sin^ X cos^ X dx = J (1 — cos^ x) cos* x sin x dx = J cos* xsinx dx — J cos* xsinxdx _ cos' a; cos^x _, - 3- + -5— + ^- Illustration 2. I cos' X dx = 1(1— sin^ x) cos x dx = I cos X dx — j sin* x cos x dx sin'x , _ = sin X — — o r C. It is seen that the process consists in combining one of the func- tions sin X or cos x with dx to form the differential of — cos x or of sin X, respectively, and of expressing the remaining factors of the function to be integrated in terms of cos x or sin z, respectively. 13 4. I sin* X dx 6. I \/sin X cos' x dx. 194 CALCULUS [§103 Exercises 1. fsin'zdx. f cos»a; •7. 7. -y== dx. 2. j sin" X COS* z dr. J ^^^'^ ^ 3. I cos* X sin' x dx. g I ^'"' ^ ^^^ r . . . t/ (cos^)' 9. I sin* a cos' a da. 6. fcos* X sin' x dx. 10- J cos^ (2x+3) sin' (2x+3) dx. (b) I sin" z COS" x dx when w and n are both even positive integers- In this case make use of the relations: sin'^ X = 5(1 — cos 2.t). cos^ X = 5(1 + cos 2x). sin X cos X = ^ sin 2x. Illustration 1. I sin' X dx = H(l ~ cos 2x) dx = \\dx — \\ cos 2x dx. _ X sin 2x ~2 r^ + ^- Illustration 2. I sin' X cos' X dx = 5 j sin' 2x dx = | 1 (1 — cos 4x) dx _ X sin 4x ~ 8 32~ "^ Illustration 3. J cos* X dx = J I (1 + cos 2x)' dx = ij(l 4- 2 cos 2x + cos' 2x) dx = ix + 4 sin 2x + I I (1 + cos 4x) dx = f ^ + 4 sin 2x + -^2 sin 4x + C. Illustration 4. J sin' X cos* X dx = J (sin x cos x)* cos* x dx = I J sin' 2x (1 + cos 2x) dx = iV J (1 — COS 4x) dx + I I sin' 2x cos 2x dx = iVx — Vv sin 4x + Vif sin' 2x + C. §103] INTEGRATION 195 Exercises 1. js'm'^xdx. 4. jsm*xdx. 2. J cos* 2x dx. 5. fsin* 3x dx. 3. / sin* X cos'' x dx. 6. j cos* 5a; dx. (c) I tan" a; rfx and I cot" x da;. Illustration 1. J tan^x dx = J tan^x (sec^x — 1) dx = „ — ftan^ x dx tan' X f/ « . V , = — 5 I (sec'^x — 1) dx tan^x , , /^ = — o tan X -r X + C. Illustration 2. J cot' X dx = J (csc^ X — ly cot X dx = I esc* X cot X dx — 21 csc^ x cot x dx + I cot x dx = — J CSC* X + csc^ X + log sin x -\- C. (d) I sec X dx and J csc" x dx, n an even integer. Illustration 1. J sec* X dx = J (1 + tan^ x) sec'^ x dx = tan x + | tan' x + C. When n is odd this method fails. (See §106.) (c) / tan" z sec" x dx and J cot*" x csc" x dx when n is a positive even integer, or when m and n are both odd. Illustration 1. J tan* X sec* x dx = J tan* x(l + tan'^ x) sec^ x dx = I tan^x + 7 tan^x + C. Illustration 2. J tan' X sec' x dx = j tan'^ x sec^ x sec x tan x dx = I (sec'^ X — 1) sec^ x sec x tan x dx = I (sec* X — sec' x) sec x tan x dx = i sec' X — 5 sec' x + C. 196 CALCULUS I §104 If m is even and n is odd the methods of §106 must be used, for the integral reduces in this case to the integral of odd powers of the secant. Exercises 9. I tan^ X sec' x dx. LO. I tan' X sec* x dx. 1. I tan* X dx. 2. 1 CSC* X dx. 3. 1 tan^ X sec' x dx. 4. I tan* X sec' x dx. 6. I cot* X dx. 6. I csc« X dx. 7. I tan* X sec'* x dx. 8. I sec* X dx. I tan 5 z sec* a; dz. 2. I (tan 2 x + tan* x) dx. 1. I sec' X tan^ x dz. 4. fctan e -f cot OY dd. 5. jtan^ede. . I sec' e tan~* 6 d9. 104. Integration of Expressions Containing s/a.^ — x% Va* + x*> \/x2 — a^ by Trigonometric Substitution. The methods of §103 find frequent application in the integration of expressions which result from the substitution of a trigonometric function for x in integrals containing radicals reducible to one of the forms Va* + x\ Va* - x\ or Vx* - a^. Illustration 1. I \/a^ — x^ dx. Let x = a sin 6. Then dx = a cos d0, and JVa^ - a;2 dx = Ja^ cos^ dO = la^{e + | sin 2d) + C = ^a2(0 + sip cog ^) + c = la' [sin-i ^ + ^, aA^":^] + C = W sin-' ^ + JxVa* - x^ + C. Illustration 2. j -\/a2 -|- a;2 ^3 dx. Let x = a tan 0. Then J Va» + x^x'dx =a5 J tan^ sec^ d9 = a'^Jtan^ 6 sec^ tan sec d^ = a»J(sec* d - sec* 0) tan sec & dd §104] INTEGRATION 197 = a6(i sec* d - I sec3 6) + C a' 1 + 1 + Illustration 3. / 5 3 5 3 + (7. dx. Let X = a sec 6. Then dx = a sec d tan d0, and sec5tan2 0d0 sec = a Jtan2 6 dd = a (tan 9 - 0) + C = a-* /-; — 1 — a sec"^ - + (7 = Vx^ — a2 — a sec-i - + C. o The integration can also be performed directly if the numerator is rationalized. Thus, rVx2 - a^ , (x2 — a^)dx ■ r-7^£= - a^ f dx = \/x2 — a2 — a sec-i - + C. a The substitutions used in these illustrations are summarized in the following table: Radical Substitution Radical becomes X = a sin ^ X = a tan 9 X = a sec d a cos 9 a sec 9 a tan 9 Va' +x^ Vx^ - a^ 198 CALCULUS [§104 Expressions involving s/ax^ + 6x + c can frequently be inte- grated by completing the square under the radical sign and making a trigonometric substitution. Illustration 1. r xdx r xdx J V3 + 2x- x2 J V'4 - (x - 1 Let X - 1 = 2 sin 9. Then a; = 1 + 2 sin ^ and dx = 2 cos 6 dd. Hence / x dx r VS + 2x - a;2 "" ^ J (1 + 2 sin 6) cos Ode 2 cos = J(l + 2 sin 0)d0 = - 2 cos ^ + C X — 1 = sin-» —2 VS + 2x - x^ + C. Illustration 2. 0^]' / rfx r rfx__ \/(2aa; - x^y ~ J [a^ - (x - a) Let X — a = a sin ^. Then x = a(l + sin d) and dx = a cos rf5. dx r a cos 5 •\/"(2ax - x^y ~~ J a'coss = ^Jsec^ede = -„ tan + C ^ 1 sin g a2 cos "^ ^ X — a 1 ~^ a2 1 , _ - V2aa; - x^ + C o'' \/2ox - x2 §105] INTEGRATION 199 Exercises dx dx 1 r dx ^ r dx r 2./>^'... »./;vfc, »-/a-.,vr^ 3./xvrTT..x. 6.J^^7^,- »./^^;r~r» 10. J (a^ - x^y-dx = 3a^f cos* sm^O de. Hint. Let S 3 . xs — a^ sin'' e, X = a sin^ 0. 4)5 dx (16 - x^)* dx r x^dx r dx r dx ' J Va^~^^' ' J xWx^^^' ^- J a;(x» - 12. fV9 - 5x2 dx. 14. f{9 ~x^)^dx. 16. 1 J (x2 + 6x + 25) 2 J (x2 + 4x - 5)^ Vl^^^" 20. Jv'2 + 6x-x^cix. 105. Change of Limits of Integration. In working the pre- ceding exercises by substitution it was necessary to express the result of integration in terms of the original variable. In the case of definite integrals this last transformation can be avoided by changing the limits of integration. L. I x^Vo^ Illustration 1. | x^ Vo^ — x^dx. Let x = a sin 6. Then dx= a cos d dd. When X = 0, sin = and ^ = 0. TT When X = a, sin ^ = 1 and ^ = o* 200 CALCULUS [§105 As X varies continuously from to o, varies continuously from TT to ^ • Hence we have I x^y/a^ - x^dz = a^\ "sin^ Q cos^ Q Jo Jo dd = a*{ld - A- sin Ad) 2 n wa* ~ 16 r** x' dx Illustration 2. 1 — -. Let x = a tan 6. Then Jo Vfl^ + x^ dx = a sec^ 6 dd. When X = 0, tan = and 6 = 0. ■r,^, . /> - 1 /. 'T When X = a, tan = 1 and ^ = 7* As X varies continuously from to a, varies continuously from IT to T- Hence we have 4 J^* x'dx C* \~* -^=== = a' 1 tan» d sec d dd = a^i sec^ d - sec d)\ VS^T^^ Jo lo = ia\2 - V2). \/a2 — x^ dx. By using the substitution X = a sin we obtain /*2 d5 JVa^ - x2 dx = a^ j cos^ «/0 = iaM^ + ^ sin 20)r Tra'' l2 ~ 4 * The above integral is of frequent occurrence in the application of the calculus. The integrand, \/a^ — x^, is represented graphic- §106] INTEGRATION 201 ally by the ordinates of a circle of radius a, center at the origin. The integral then represents the area of one-quarter of this circle. (See §§64 and 65.) The value of any integral of this form may be written down at once. Thus, I V4 -{x- 5)2 dx = I \/4 - M^ rfw = Jo Jo 7r22 s: ^^TT^ _ _ 7r(3-hz^) \/3 + z^ - a;2 dx = -. Exercises 1. I (9 - a;*)* dx. 7. I Jo Jo (o^ + X ,. f— i^. s. r Jo V2ax-x' J^A r x^dx g r_ • X V9^r^ ' Jo (^ ,. r dx . ^, r Vi" X (x^ - 4)' dx + 16)2 dx x'^ix^ - 9)^ 6 I ^25 - x^dx. 11. I \/9 - (x - 4)2 dx. V^9 _ a;2 dx. 12. I Vb'' - 2/2 _ x2 dx. Jo 106. Integration by Parts. The differential of the product of two functions u and v is d{uv) = udv ■}- V du. (1) Integrating we obtain uv = \udv -\- \vdu From which r u dv = uv — J V du. (2) This equation is known as the formula for integration by parts. It makes the integration of u dv depend upon the integration of d» and of v du. 202 CALCULUS [§107 Illustration 1. j x log x dx. Let log x = u and x dx = dv. The application of (2) gives I xlogx dx = Ix^ log X — I I x^- dx = |x2 log X - lx^ + C. Illustration 2. I xe'*da;. Let e^'dx = dv and x = u. The application of (2) gives I xe^'dx = \xe^' — \ \ e^' dx = |xe3' - W + C = ie3'(3x - 1) + C. If we had let xdx = dv and e'* = w we should have obtained a more complicated expression to integrate than that with which we started. Exercises 1. I x* log X dx. 2. j X cos x dx. 3. I sin~i x dx 4. I x'' e*' dx. (Apply formula (2) twice in succession.) 6. I tan~i X dx. 9. I x sin' x dx . 6. I X sin X dx. 10. I log x dx. 7. I x" log X dx. 11. I x^ sin 2x dx. 8. I x'^ tan~^ 2x dx. 12. I sin x log cos x dx. 107. The Integrals J e" sin nx dx, J e** cos nx dx. Let M = sin nx and dv = €"dx. Then Jc* sin nx dx = -e»* sin nx ( e"' cos nx dx. a a J A second integration by parts with u = cos nx and dy = e"dx gives f . , 1 . n w* f . . I e"sm nxdx = -e<»*sin nx ;e"cos nx x I e" sm nx dx. J a a^ a^ J The last term is equal to the integral in the first member multiplied §107] INTEGRATION 203 by -^- On transposing this term to the first member we obtain ^ — I e"^ sm nx ax = ^ (a sm nx — n cos nx) + C. Then /©ax e sin nx dx = —^-r — i (a sin nx — n cos nx) + C eax sin (nx - a) + C, (1) where and cos a = sm a Va^ + n2 q The student will show in a similar way that /gax e"cos nx dx = ~ir~, — o (n sin nx + a cos nx) + C a2 _i- n2 V Va^ + n2 where a^ + n^ eaz cos (nx - a) + C, (2) cos a = and n sm a = Va^ Exercises The student will work exercises 1-5 by the method used in obtaining (1) and (2) above. In the remaining exercises he may obtain the results by substituting in (1) and (2) as formulas. 1. fe-" sin 7t dt. 6. \e-'^ cos bt dt. 2. fe-^ cos St dt. 7. je-"-" sin ut dt. 3. fe-o" sin St dt. 8. fc-o-^ cos co< dt. 4. fe-o-^' cos 4< d<. 9. fe-"'" cos 5t dt. 5. je~''smxdx. 10. j e'"-' s'm 4:t dt. 11. Find a in exorcises 1-10. 204 CALCULUS [§109 108. J sec^ X dx. This integral can be evaluated by a method similar to that used in the last article. /sec...x=/sec.sec..<*. = sec X tan x — \ sec x tan^ x dx. Since tan^ x = sec'^ a; — 1, I sec' xdx = sec x tan ^ — j sec' x dx + j sec x dx. Transposing the next to the last term to the first member, dividing by 2, and integrating the last term we have I sec' X dx = i [sec x tan x + log (sec x + tan x)] + C. Exercises 1. I csc^ X dx. 5. j VoM-^ rfx. 2. jsec" X dx. 6. J Vx^ — 4x + 11 (ia;. ,. dx. 7. I , dx. Va^ + x2 dx. 8. I \/x2 — 9 dx. t/3 109. Wallis' Formulas. Formulas will now be derived which make it possible to write down at once the values of the definite integrals: I sin" d de, 2 COS" 6 dd, and f I sm« 9 cos" 6 dd, where m and n are positive integers greater than 1. J sin" Odd = \ sin"-i sin d dd. Jo §109] INTEGRATION Integration by parts gives 205 sin" ddd = — sin»-^ d cos B a + (n - 1 ) I sin"-2 e cos2 B dB 2 0(1 - sin2 B) dB = (n — 1) I sin""' = (n - 1) I sin»-2 BdB - (n - 1) \ sin» Jo Jo g the last term and dividing by n J sin" BdB= ^ ~ I sin"-^ B « Jo rfe. On transposing the last term and dividing by n we obtain dB. This equation can be regarded as a reduction formula for expressing r sin" B dB in terms of an integral in which sin B occurs with its exponent diminished by 2. Applying this formula successively we obtain I sin" BdB = 1 I ' sin"-" B dB Jo n n- 2X n — In — 3n — 5 n 71 — 2 n — 4 (n-l)(n-3) ■^BdB n{n - 2) (n-l)(n-3) -^ ( sin BdB'iin is odd. t n{n - 2) • [(n-l)(n-3) 4-2 ^r dB sin" dO 4-2 n(n - 2) • • • 31 (n-l)(n-3) • • • S-Itt n(n - 2) • • • 4 2 2 if n is even. if n is odd. if n is even. 206 CALCULUS [§109 From the fact that the integrals f- sin" X dx Jo and ''2 £ COS" X dx represent the areas under the curves y = sin" x and y = cos" x , IT respectively, between the limits x = and x = ^, it is clear from the graphs that I COS" xdx = I Jo Jo COS" xdx = \ sin" x dx. The results obtained can be expressed in the single formula f COS. ... ./W... - ^^^i^^^^^,-^^..) where a = 1 if n is odd, and a = ^ if n is even. In a similar way we shall evaluate T f: sin" d COS" 6 dd. I 2 ri sin" d COS" d dd = | sin^-i ^ cos" d sin ^ f/0 sin*"-^ d cos"+^ 5 m — _ . + ^ , 1 I sm'"-^ d cos"+2 d0 n + 1 X ^ I ^ I sm^-2 COS" 0(1 - sm2 d) dd 1/^2 m — 1 P J I sm^-^ecos"^^^ - ^^-pj I siw-dcos^dde. n + §109] INTEGRATION 207 Transposing the last term to the left member of the equation b-m]£ sin" 6 COS" 6 do = sin" 6 COS" 6 dd m-1 C^ . I , I SI ^ + 1 Jo m-1 r sin^-^acos"^^^ Apply this formula successively and obtain r sin" 9 cos° d dd (w — 1) (w — m — 1 m—S m -\- n m -{- n 3_ p -2J0 sm'"-2 0cos"ed^. sin"'"''^ COS" 0d^ {m -\- n) {m -\- n (m-1) (m-3) Z^pTV(^pcos«ed^ if m IS even (m + n) (m + n — 2) (m-i)(m-3)- • •i-(n-i)(n-3) ■^ n •(^ + 3)J„ sin 6 COS" ^ dd if mis odd (in+n)(mH-n— 2) (m-i)(in-3)- •(n+2)(n)(n-2) i-(n-i)(n-3)- (m+n)(m+n— 2) • • • (n+2)(n)(n— 2) (m-i)(m— 3) • • • 2 IT., . if n is 2 2 if n is odd and m is even. , w , N / , x/ ; — r if n is either even or (m+n)(m+n— 2) • • • (n+3)(n4-i) j •, j • jj ' odd, and m is odd. The right-hand member of the last formula of this group can be put in a form similar to the others by multiplying numerator and denominator by (n — l)(n — 3) • • • 2 or 1. It becomes (m-i)(m-3) • • • 2 • (n-i)(n-3) • • • 2 or i (m+n)(m+n-2) • • • (n+3)(n+i) (n-i) (n-3) • • • 2 or i IT These formulas for | sin"* ^ cos" 6 dd can all be expressed in the single formula X r sin-d cos"d dd = (m-i)(m-3)--2ori(n-i)(n-3)-2ori Jo (m + n) (m + n- 2)---2 or I «• (2) IT where a = 1 unless m and n are both even, in which case a = „* 208 CALCULUS [§109 Illustration 1. By formula (1) _ 8-6-4-2 128 r r B; sin^ Odd 9-7-5-31 315 Illustration 2. X cos*edd = ^!!: = ^TT. 4-2 2 16 Illustration 3. By formula (2), 4.0.0 1 sm^ X cos' X dx = = __. 8-6-4-2 24 Illustration 4. J 2 . 4-9'^-1 S sm^x cos^a: dx = ^ '^ = _r_. 9-7-5-31 315 Illustration 5. sin'x cos^a; dx _ 5-31-31 X 3t 10-8-6-4-2 2 512 Exercises cos" X dx. 1. I sin^ e de. 7. I Jo Jo 2. I cos^^ede. 8. I sin^<^d<^. r 3. I cos^ede. 9. I sin^ a; COS* X dx. 4. I sin'^ede. 10. I sin^ X cos" X dx. Jo Jo X X 6. I COS* e do. 11. I sin< x cos^ x dx. Jo Jo x^ !L 6. I sin«fldfl. 12. I sii Jo Jo sin' <f> cos ff> d4t. §110] INTEGRATION 209 13. I sin* X cos* x dx. 16. I x^ {a^ — x*) dx. 14. I cos^ X sin6 x dx. 17. I (a* - x^)^^^. Jo Jo 15. I (a2-x2)^dx. 18. I x(a^ - x^)^dx. Jo Jo 19. j a}{\ - cos e)HQ = 4a2 j sin^ | dd. Let 0' = 2- Then d0 = 2de' and «' = | wnen fl = tt, and fl' = when = 0. Hence a}\ (1 - coseyde = 8a2 ( sin^O'dO'. Jo • Jo WaUis' formula can now be applied. By transformations similar to the foregoing many integrals can be put into a form to which Wallis' formulas can be applied. 20. I cos* 2ede = \ I cos* e' do'. Jo Jo r _ 22. I x\/2ax - x* dx. 21. I (2ax - x*)^dx. (Substitute x = 2a sin* 6.) I »2a fl. sixi X I b cos X 110. Integration of f — . — ^— dx. Integrals of this ' c sm X + d cos X X form can be reduced by the substitution, z = tan g. In making this substitution it is necessary to express sin x, cos x, and dx in terms of z. This is easily done as follows. (The student is ad- vised to observe the method carefully, but not to learn the results as he can readily obtain them whenever needed.) Since z = tan ^> X = 2 tan~* z, 14 210 CALCULUS [§110 and dz dx = 2~. — j — 7, Further, and Then and COS;^ = 2 X \ X V'l+2=' sec 2 -y/l + tan2 o . X ^ X X sin 7. = tan 7=. cos ^ 2 2 ^"-^ 2 VlT „ . a; a; 2z sin X = 2 sin ^ cos ^ = 2 2 1+^2 X X 1 - z^ cos a; = cos^ 7^ — sin^ - = '■/r 2 2 1 + z2 dx Illustration 1. I ^ — r~j On making the substitution ' ^ + 4 cos X X z = tan ^ we obtain by using the values just found for cos x and dx in terms of z,^ 2dz ! l-\-z' ^ 2 ' ^^ 1 + 4, , _3 ^/r 1 -z^ J 1 +2^ + 4(1 -z^) 1 +2^ dz -/^ 322 2 C \/3dz ■J- VsJ 3z2 - 5 2 ,„gV^i^- + c 2\/3 \/5 \/3z + VS V3 tan I + VS = A \/l5 log ^ 3 + C. \/3 tan 2 — \/5 1 The student will derive these values in each problem worked in order to famil- iarize hinist'lf with the method. §111] INTEGRATION 211 Illustration 2. | _ #:7— • Let z = tan % Then 2dz dz f dx ^ r i+g' _ ^ r j5-3sinx I Qz ~ ^J 5 + 522 - 62 1+02 dz . f dz _ , r V .2 _ e, + 1 - J _ (2-|)^ + ii = f • I tan-i -^ + C = ^ tan-i ^^^ + C' 5 tan 2 ~ 3 = § tan-i -. + C. Exercises The student will find cos x, sin x, and dx in terms of the new vari- able in each of the exercises. . r dx r ^' J 3 + 5 cos x' ^' J 1 i. C l^- 7 fs J 5 — 3 cos X J 1 , r ^-^ . 8 ' r ^ J 4 — 5 sin a; * J sin a / sin xdx r • J 2 + sin x' ^- J : /COS iC 1 o I o 1^^- 10. I 3 + 2 cos T I . — 3 sin X + 4 sin x + 2 sin X + sin X dx dx. dx. x(l + cos x) dx 3 sin 2x dx 4 — 5 cos 2x 111. Partial Fractions. A rational fraction is the quotient of two polynomials, e.g., ttox™ + flix"*-^ + • • + am-ix + a„ 4>{x) box" + bix"-^ + • • • + b„-ix + 6„ /(x) (1) 1 The integrand is not in the form given in the heading of this article, but the sub- stitution z = tan 2 enables us to transform any expression containing only integral powers of sin x and cos x into a rational function of «, i.e., into a function containing only integral powers of z. 212 CALCULUS [§111 If the degree of the numerator, m, is greater than or equal to the degree of the denominator, n, the fraction can be transformed by division into the sum of a polynomial and a fraction whose nu- merator is of lower degree than the denominator. In this case the division is always to be performed before applying the methods of this section. The integration of a rational fraction cannot in general be aC' complished by the methods which have been given if the degree of the denominator is greater than 2. Illustrations will now be given of a process by which a rational fraction can be expressed as the sum of fractions whose denominators are either of the first or second degrees. Illustration 1. x2 + 2 /. dx. -2x2-9x4-18 Factoring the denominator X' - 2x2 - 9x + 18 = (x - 2)(x - 3)(x + 3). Assume x2 + 2 A ,_B_ C x3-2x2-9x + 18 X -2^x -3"^x4-3' where A, B and C are to be so determined that this equation shall be satisfied for all values of x. Clearing of fractions x^ + 2 = Ax2 - 9A + 5x2 + Bx - QB + Cx^ - 5Cx + 6C = {A+ B + C)x2 + (B - 5C)x - 9A - 65 + 6C. On equating the coefficients^ of x^, x, x", we obtain the following three equations for the determination oi A, B and C. A-hB + C = 1. 5 - 5C = 0. - 9A - 6B 4- 6C = 2. 1 In applying this process use is made of the fact that if two polynomials in x are identically equal, the coefBcients of like powers of x are equal. Thus, given the identity a«" + aia;"-l + • • • + ccn-i x + a„ = /Sox" + /3ii"-l + • • • +0n-i x + P„, then ao = Po ai = /Si §111] INTEGRATION 213 From these equations A= -h 5 = V. C = U. Hence x^ + 2 -6 11 11 a;3 _ 2x2 _ 9a; _|_ 18 5(3. _ 2) ' 6(x - 3) ^ 30(a; + 3) and r x^+2 ^ r_rfx J a;3 _ 2x2 - Qx + 18 ^ ^J X- = - l\og{x-2) + V log (x-3)+U log (x + 3) + C. 2 + V Tr-. + U r^ Short Method. The foregoing method of determining the values oi A, B, • • •, by equating coefficients of like powers of X, is perfectly general. However, a shorter method can sometimes be used. Thus in the illustration just given write the result of clearing of fractions in the form a;2 + 2 =A(x- S)ix + 3) + 5(x - 2)(a; + 3) + C(x -2)(x- 3). Since this relation is true for all values of x, it is true f or x = 2. On setting x = 2, we obtain 6 = - 5A. Hence A = -i On setting x = 3^ we obtain 11 = 6B. Hence B = V. On setting x = — 3, we obtain 11 = 30(7. Hence /n* _ -Li Illustration 2. x2 + l / (x + lKx-l)'^''- 214 CALCULUS [§111 Let x' + l _ A B C D (x + l){x - 1)3 ~ X + 1 "•" (x - 1)3 + (x - 1)2 "^ a; - 1* On clearing of fractions, x^ + I = Aix - ly + Bix + 1) +C{x-l){x-{-l)+D(x-iy{x+l), or x^ + 1 = Ax^ - 3Ax2 + 3Ax- A+ Bx + B + Cx^ - C + Dx^ - Dx^ - Dx -j- D. In the first form put x = 1. Then 5 = 1. In the first form put x = — 1. Then - 8A = 2. Hence A= -h Equating coefficients of x' in the second form A + D = 0. Hence D = - A = I Equating coefficients of x^ in the second form, -SA + C - D = I. Hence O— 14-1-4 — 2' Consequently x^-\-l ^ -1 1 1 1 (x + l)(x - ly 4(x + 1) "*" (x - ly "^ 2(x - 1)2 ■+" 4(x - 1) and J (X + 1)(X-1)3'''' ^Ja^ + l+J (.r_l)3 = _ ^ log (X + 1 ) - 2^^^, - 2(^ + i ^ §1111 INTEGRATION 215 Illustration 3. /i 3.r2 - 2a; + 2 ax. {x - l)(a:2-4x + 13) Let 3x2 -2a; + 2 ^ 5.r + C (x -l)(a:2-4x + 13) x-1 ' x^ - 4a; + 13 Clearing of fractions, 3x2 - 2x + 2 = A{x^ - 4x + 13) + Bx(x - 1) + C(x - 1), or 3x2 - 2x + 2 = Ax^ - 4Ax + 13A + 5x2 ^ Bx + Cx - C. In the first form put x = 1. We obtain 3 = lOA. Hence A = A. Equating the constant terms in the second form, 13A - C = 2. Hence 18 - C = 2 and n — i-9 O — 1 u- Equating the coefficients of x^ in the second form, A+ B = 3. Hence 5 = 3 — "i = To . Consequently C 3x2-2x + 2 . 3 r rf^ , 1 r 27x + 19 J(x-l)(x2-4x + 13)^^ = '^"J^^ + "^"Jx-2-4x + 13''^ = A log (x-1) + ,ij ^ ._4^^i3 + i3 J (^^ dx -2)2 + 9 X — 2 = -h log (x-1) + 2-& log (x2-4x+13) + H tan-i -3— + C Illustration 4. 2x dx (r+x)(l+x2)5 Ji 216 CALCULUS [§111 Let 2a; ^ Bx -\- C Dx + E (1 + x)(l + x^y ~ 1 + a; "•" (1 + x^y "^ (1 + x-")' In Illustrations 1 to 4 a fraction was broken up into "partial fractions." The denominators were the factors of the denomina- tor of the given fraction. In Illustrations 1 and 2 the factors were all real linear factors, while in Illustrations 3 and 4 there were also factors of the second degree which could not be factored into two real linear factors. The method of procedure will be further indi- cated by the following examples. They will be grouped under the numbers I, II, III, and IV, corresponding to Illustrations 1, 2, 3, and 4. I. Factors of denominator linear, none repeated. , . x' + 5 ^ g C ^"^ {x - l)(x + l)(a; -3) a;-l"'"x + la;-3 ., . x^ -I- 2a: + 7 A_ B ^' (a; + 4) (2a; 4- 3) (x- 2) (3a; + 1) a; + 4 "*" + ^ a; - 2 ' 3a; + 1 II. Factors of denominator linear, some repeated. a;^ + 2a; + 5 ^_ . ^_ . C' (x-2)2(a;-3)'(x + l) (a; - 2)^ ' x-2 ' {x-ZY D E (a; -3)2 ' a; -3 ' x + 1 ... x» + 4a; - 2 A B (2x -f l)8(a; + 3)(a; - 4)^ (2a; + l)^ ^ (2a; + 1)^ 2x + 1 ' x + S ' (x - 4)2 ' a; - 4 III. Denominator contains factors of second degree, none repeated. x2 + 7x + 3 _ Ax + B C ^^' (x2 + 4) (a; - 2) ~ x* + 4 + x - 2* „ . x» - 3x + 5 Ax-\-B Cx-^D "' /•-V.2 _i_ o^/•^2 _ /i^ j^ '7\/'^ 1 o^ "~ ™2 I o I (x2 + 2)(x2 - 4x + 7)(x + 3) x2 + 2 ^ x2 - 4x + 7 + _^. X -\- 3 x2 + 2x - 5 _ Ax + B C__ D ^^^ (x2 + 7)(x-2)2 x2 + 7 "^ (x-2)2 + x -2' §111] INTEGRATION 217 IV. Denominator contains factors of second degree, some repeated. a;3 + 2x2 + 5 Ax + B (a) (x2 + 2x + 10)2(x2 + 3)(a; + 2) (x^ + 2x + 10) ^ Cx + D Ex-^F G a;2 + 2x + 10 ' x^ + S ' x + 2 Exercises X3 + X - 10 '^*' *• J (i+1 r (x-4)^x r J X' - 6x« + 9x °' J r x« + x^ + 7x + 1 dx KxM-T)' (3 + 4x - x") dx (x- lKx''-2x + 5)' 5x2 _|_ 13a; _ 7 (x+4)(2x+l)2 dx. dx I* + x3+3 — 9x dx. (Divide numerator by denominator.) CHAPTER XIII APPLICATIONS OF THE PROCESS OF INTEGRATION. IMPROPER INTEGRALS 112. In this section a brief summary and review of the appHca- tions of the process of integration will be given. 1. Area under a Plane Curve: Rectangular Coordinates. nb A = I f{x)dx. See §64, and Fig. 46. 2. Area: Polar Coordinates. -'£ See §99, and Fig. 74. 3. Length of Arc of a Plane Curve: Rectangular Coordinates. dx i:>F(i)'*- See §69, and Fig. 49. 4. Length of Arc: Polar Coordinates. See §98, and Fig. 72. 5. Volume of a Solid of Revolution. »b F = I Try"^ dx See §68, and Fig. 49. 218 §112] IMPROPER INTEGRALS 219 6. Surface of a Solid of Revolution. 'x = b ds = 2ir I y Jx =a y ds. See §70, and Fig. 49. 7. Water Pressure on a Vertical Surface. P = k\ uz du, Ja where z denotes the width of the surface at depth u and k = 62.5 pounds per cubic foot if u and z are expressed in feet. See §72, and Fig. 50. 8. Work Done by a Variable Force. See §67. Exercises 1. Find the area in^the first quadrant between the circle x^ + y'^ = a* and the coordinate axes. The definite integral which occurs in the solution of this problem is of very frequent occurrence. See Illustration 3, §105. 2. Find the area bounded by the lemniscate, p^ = a^ cos 20. 3. Find the length of one quadrant of the circle x^ + 2/^ = o*, or X = a cos 0, y = a sin 0. 4. Find the length of p = 10 cos 0. 6. Find the volume of a sphere of radius a. 6. A solid is generated by a variable square moving with its center on, and with its plane perpendicular to, a straight line. The side of this square varies as the distance, x, of its center from a fixed point on the line, and is equal to 2 when a; = 3. Find the volume generated by the square when its center moves from a; = 2 to x = 7. 7. Find the area of the surface of a sphere of radius a. 8. The unstretched length of a spring is 25 inches. Find the work done in stretching it from a length of 27 inches to a length oi 220 CALCULUS (§112 29 inches, if a force of 400 pounds is necessary to stretch it to a length of 26 inches. 9. A trough 3 feet deep and 2 feet wide at the top has a parabolic cross section. Find the pressure on one end when the trough is filled with water. 3 3 2 10. Find the length of the curve x* +2/' =a^, orx = ocos' 6, y = a sin' B. 11. Show that the work done by the pressure of a gas in expanding from a volume i^i to a volume V2 is given by I p dv. Jn where p is the pressure per unit area. Hint. Take a cylinder closed by a piston of area A forced out a distance Ax by the expanding gas. Denote by At/; the work done by the gas in expanding from a volume r to a volume v -\- Av, Then, I] p dv. -f 12. Find the area of one quadrant of the ellipse x = o cos 6, y = b sin 6. 13. Find the area of one loop of the curve p = a cos 26. 14. Find the length of the cardioid, p = a(l — cos 6). 15. Find the volume of the ellipsoid of revolution generated by revolving the ellipse "i + j^i ~ ^ about the X-axis; about the F-axis. 16. A volume is generated by a variable equilateral triangle moving with its plane perpendicular to the JC-axis. Find the volume of the solid between the planes x = and x = 2, if a side of the triangle is equal to 2x^. 17. Find the area of the surface generated by revolving about the X-axis the portion of the arc of the catenary a ^=2 + e ■■] between (0, a) and (xi, yi). 18. Find the area under one arch of the cycloid x = a {d — sin d), y = a(l — cos 9). §112] IMPROPER INTEGRALS 221 19. Find the length of that portion of 9y^ = x' above the Z-axis between x = and x = 3. 20. Find the volume generated by revolving the portion of the catenary y « r - . --1 2 e" +e « between x = and x = b about the X-axis; about the F-axis. 21. Find the volume generated by revolving the hypocyoloid 2 3 1 x^ + y^ = o*, or X = a cos^ 0, y = asm^ 0, about the X-axis. 22. Find the area included between the parabolas 47/^ = 25x and 5x^ = \oy. 23. Find the area between the X-axis, the curve y = x'' — 4x + 9, and the ordinates x = 1 and x = 7. 24. Find the area between the curve y = sin x, the X-axis, and X = and x = w. 26. If a gas is expanding in accordance with Boyle's law, -pv = C, find the work done in expanding from a volume vi to a volume V2. Represent the work graphically by an area. 26. Find the work done if the gas is expanding in accordance with the adiabatic law, py* = C. Hint. From the result of Exercise 11, C k A-k _ „.l-ft Vi^-'). Now, C == PiVi* = ^2^2*. Hence W = j-iri; (P2i'2 - PlVl). Represent the work graphically by an area. Use the same scale as in Exercise 25. x^ V^ 27. Find the area of one quadrant of the ellipse Ta + g" ~ ■'■• ^^ Exercise 1. 28. Find the length of p = e"* from ^ = to ^ = 27r. 29. Find the length of p = e""^ from ^ = to (? = - =" , if a is assumed positive. 30. Find the area bounded by the cardioid p = a(l + cosO). 31. Find the area bounded by p = 10 sin d. 32. Find the area bounded by the hypocycloid x = o cos^ 9, y = a sin^ d. 222 CALCULUS [§112 33. Find the area between y^ = Ax and y^ = 8x — x'^. 34. Find the work done by a gas in expanding isothermally from an initial volume of 2 cubic feet and pressure of 7000 pounds per square foot to a volume of 4 cubic feet. 36. Find the work done if the gas expands adiabatically. Take k = \h the value for steam. (See Exercise 26.) 36. Find the pressure on a trapezoidal gate closing a channel con- taining water, the upper and lower bases of the wet surface being 25 feet and 18 feet, respectively, and the distance between them being 10 feet. 37, Find the area between the catenary |[^ea+e aj, the X-axis, and the ordinates x = and x = a. 38. Find the length of p = ad from = to fl = 27r. 39. Set up the integral representing the length of one quadrant of the ellipse x = a cos 0, y = b sin 0. 40. Find the volume generated by a circle of variable radius mov- ing with its plane perpendicular to the X-axis, between the planes X = 2 and x = 8. The radius is proportional to x^ and is equal to 54 when X = 3. 41. Find the volume generated by revolving one arch of the cycloid X = a{d — sin d), y = a(l — cos 6) about the X-axis; about the tangent at the vertex. 42. Find the area of the surface generated by revolving a quadrant of a circle about a tangent at one extremity. 43. If the density of a right circular C3dinder varies as the distance from one base, find the mass of the cylinder. if the altitude is h and the radius of the base is r. 44. The force required to stretch a bar by an amount s is given by „ Eos where E is the modulus of elasticity of the material of the bar, o is the area of the cross section, and L is the original length. Find the work that is done in stretching a bar whose unstretched length is 400 inches to a length of 401 inches, if £ = 30,000,000 pounds per square inch and o = 1.5 square inches. 46. Find the area of one loop of p = 10 sin Zd. §113] IMPROPER INTEGRALS 223 46. Find the length of y = ^\ |[^e- +e~aj from (0, a) to {xi, y,). 47. Find the length of one arch of the cycloid x = a{d - sin. e) , y = a{l — cos e). 48. Find the volume of the anchor ring generated by revolving the circle x^ + (y - by = a^ about the X-axis, a being less than b. a 49. Find the area of the small loop of p = a sin^ ^ * o 60. Find the work done in pumping the water out of a cistern 20 feet deep, in which the water stands 8 feet deep, if the cistern is a parabo- loid of revolution and the diameter at the surface of the earth is 8 feet. 51. Find the volume included between two equal right circular cylinders, radius a, whose axes intersect at right angles. 62. Find the area of the surface generated by revolving one arch of the cycloid x = a{e — sm 6), y = a{l — cos 9), about the X-axis; about a tangent at the vertex. 63. Find the area bounded by p =3+2 cos 6. 64. Find the area bounded by the small loop of p =2 + 3 cos B. 66. Find the area of the surface generated by revolving the cardioid p = a(l + cos 6) about the polar axis. 56. Find the volume bounded by the surface of Exercise 55. 113. Improper Integrals. Since — , becomes infinite at ■s/x — 1 X = 1, the definite integral r 1 dx : ■s/x must not be evaluated by the usual process. For, the assumption has been made that in the integral fix) dx fix) is a continuous finite function at x = a and x = 6 as well as at all intermediate points, and the evaluation of this integral was based on the area under the curve y = fix). In this case 224 CALCULUS [§113 becomes infinite at the lower limit. The area under the curve 1 y Vx^^ between the ordinates x = 1 and x = 7 has no meaning. In fact the integral in question has no meaning in accordance with the definition of a definite integral already given. A new definition is necessary. We define •^ 1 dx as r im / dx, +, Vx - 1 where r/ is a positive number, if this limit exists. Otherwise the integral has no meaning. Now, 1™ f -yL^ dx - >s, (2Vi^i) r Ji+, y/x-\ I14-, = I'i? (2 V6 - 2v^) = 2V6. Since the limit exists we say that J 7 1 ■ dx = 2\/6. Graphically this means the limit as 17 approaches zero of the area under the curve y = , between the ordinates x = 1 + ?; V a; — 1 and X = 7, exists and is equal to 2\/6. Exercise 1, Show that i dx ], {X - 1)' exists if < w < \. On the other hand, when n = 1, im r 1 , lim , / ^sV Ji+v X I ^ 1^ 1+17 §113] IMPROPER INTEGRALS 225 This limit does not exist and consequently we say that 1 / x-1 dx has no meaning or does not exist. Graphically this means that the area under the curve y = 1 X- 1 between the ordinates x = 1 + t] and x = 7 increases without limit as rj approaches zero. Exercise 2. Show that S. dx X (^ - 1)" does not exist if n ^ 1. (Note that the case n = 1 has just been considered.) If n<0 no question as to the meaning of the integral can arise. Why? A definite integral in which the function to be integrated becomes infinite at the upper limit is treated in the same way. Thus I dx is defined as dx lim p"" vn=- where ?; is a positive number, if tliis limit exists. Exercise 3. Show that i dx (1-^)" has a meaning in accordance with this definition if < n < 1, and that it has no meaning if n > 1. If » < no question can arise as to the meaning of the integral. It is easy to see how to proceed in case the function under the 15 226 CALCULUS [§114 integral sign becomes infinite at a point within the interval of integration. Thus dx ;;» where n is a positive integer, I is defined as limrr-" dx__ r dx where r; is a positive number, if this limit exists. If not, the integral has no meaning. Tf n < no limit process is necessary. Exercises Evaluate the following integrals if they have a meaning: dx 3x-4 dx 2 I 3 dx 2)^ * Jo ^'' ■ Jo VoT^' • J_, X Jo "^ X V^^=^^ Jo (X- 4 f^^ dx g p dx 12. Find the area between the curve y' = ^ — — — > its asymptote and the X-axis. 114. Improper Integrals : Infinite Limits. In §113, the interval of integration was finite. In other words neither of the limits oi the integral fmdx was infinite. The integral p dx Jo ^' + «^ will be defined as lim r dx 6= CO 1 3-2 1 ^ §114] IMPROPER INTEGRALS 227 if this limit exists. Now lim b = lim 1 , _,b 1 TT 0-" a a a 2 im r d^ ^ ^^"^-tan-i- 6™ I 2 I — 2 represents graphically the limit of the area under the curve y = — ;rn — ; between the ordinatcs x = and x = h " x^ + a^ as & increases indefinitely. Consider dx X ' lim / dx lim lim , , inci I X But log 6 increases without limit as h increases witnout limit. Hence | — has no meaning. Exercises Evaluate the following integrals if they have a meaning: 2. I e-'dx. 4. I x'^e-'dx. Jo Jo 6. Find the area between the witch, ij = VXT^' ^^^ ^^® ^^^^ °^ ^* CHAPTER XIV SOLID GEOMETRY 115. Coordinate Axes. Coordinate Planes. Just as the posi- tion of a point in a plane is given by two coordinates, for example by its perpendicular distances from two mutually perpendicular coordinate axes, the position of a point in space is given by three coordinates, for example by its perpendicular distances from three mutually perpendicular planes of reference, called the coordinate planes. Let the three coordinate planes be those represented in Fig. 75, viz., XOY, called the ZF-plane, YOZ, called the FZ-plane, and ZOX, called the ZX-plane. Jhen the position of the point P whose perpendicular distances from the YZ-, ZX-, and XF-planes / Fig. 75. Fig. 76. are 2, 3, and 1, respectively, is represented by the coordinates 2, 3, and 1. The lines of intersection of the planes of reference are called the axes. Thus X'OX, Y'OY, and Z'OZ, Fig. 76, are called the axes of x, y, and z, respectively. The coordinates of a point P measured parallel to these axes are known as its x, y, and z coordinates, respectively. Thus for the particular point P of Fig. lb,x = 2,y = 3, and z = 1. More briefly we say that the point P is the point (2, 3, 1). In general, (x, y, z) is a point whose coordinates are x, y, and z. If these coordinates are given the 228 §115] SOLID GEOMETRY 229 position of the point is determined, and if a point is given these coordinates are determined. The relation between a function of a single independent variable and its argument can be represented in a plane by a curve, the ordinates of which represent the values of the function correspond- ing to the respective values of the abscissas. Thus, y = j{x) is represented by a curve. To an abscissa representing a given value of the argument there correspond one or more points on the curve whose ordinates represent the values of the function. In like manner a function of two independent variables x and y can be represented in space. Choose the system of coordinate planes of Fig. 75. Assign values to each of the independent variables X and y. These values fix a point in the XF-plane. At this point erect a perpendicular to the XF-plane, whose length z repre- sents the value of the function corresponding to the given values of the arguments. Thus a point P is determined. And for all values of X and y in a given region of the XF-plane there will, in general, correspond points in space. The locus of these points is a surface. The surface represents the relation between the function and its two independent arguments just as a curve represents the relation between a function and its single argument. Thus if 2 = ± V25 -x^ - y^ = /(x,y), ± \/l2 are the values of the function corresponding to the values x = 2 and 2/ = 3. Then th e points (2, 3, 2 \/3) and (2, 3, - 2V3) lie on the surface 2 = ± V^25 -x^- y\ If x = -3and?/ = 1, 2 = ± -s/l5- The corresponding points on the surface are (— 3, 1, -v/lS) and (-3, 1, - \/l5)- The coordinate planes divide space into eight octants. Those above the XF-plane are numbered as shown in Fig. 76. The oc- tant immediately below the first is the fifth, that below the second is the sixth, and so on. The points (2, 3, 2\/3) and (2, 3, - 2\/3) lie in the first and fifth octants, respectively. The points (-3, 1, \/l5) and ( — 3, 1, — \/l5) lie in the second and sixth octants, respectively. The locus of points satisfying the equation z = ± V25 - x^ -tj^ (1) iS a sphere of radius 5. For, this equation can be written in the 230 CALCULUS [§117 form x^ + 2/^ + 2^ = 25, which states that for any point P on the surface (1), OP = Vx"^ + y^ + z^ = 5. The left member is the square of the distance, OP, of the point P (x, y, z), from 0, since OP is the diagonal of a rectangular parallelopiped whose edges are x, y, and z. If then the coordinates of P satisfy (1), this point is at a distance 5 from the origin. It lies on the sphere, of radius 5, whose center is at the origin. 116. The Distance between Two Points. The student will show that the distance d between the two points (xi, yi, Zi) and (X2, t/2, Zz) is d = V(x, - xi^ + (i/2 - yir + (22 - zi)K (1) See Fig. 77. If the point (xi, yi, Zi) is the origin, (0, 0, 0), the expression for d becomes P = VX2^ + 2/2^ + 22= (2) Fig. 77. Fig. 78. Exercises Find the distance between the following points : 1. (1, 2, 3) and (3, 5, 7). 2. (1, -2, 5) and (3, -2, -1). 3. (0, -3, 2) and (0, 0, 0). 4. (0, 0, 3) and (0, 2, 6). 6. (0, 0,-5 and (2, 0, 6). 6. (-3, 2, -1) and (0,0, 0). 117. Direction Cosines of a Line. Let OL, Fig. 78, be any line passing through the origin. Let a, jS, and y be, respectively, the angles, less than 180°, between this Hne and the positive direc- tions of the X-, Y-, and ^-axes. These angles are called the §118] SOLID GEOMETRY 231 direction angles of the line, and their cosines are called the direction cosines of the line. Let P, whose coordinates are x, y, and z, be any point on the line. Let OP = p. Then X = p cos a, y = p cos /3, and z = p cos y. Squaring and adding the above equations we obtain 2-2 _|. y2 _^ 2^ = p2(cos^ a + cos^ /3 + cos^ 7). Since 3.2 _j_ j^2 ^ 2^ = p2, cos^a + cos^/3 + cos^Y = L (1) The direction cosines of any line are defined as the direction cosines of a parallel line passing through the origin. Then, the sum of the squares of the direction cosines of any line is equal to unity. Exercises Find the direction cosines of the lines passing through each of the following pairs of points. 1. (0, 0, 0) and (1, 1, 1). 2. (0, 0, 0) and (2, -3, 4). 3. (0, 0, 0) and (-1, 2, -3). 4. (1, 2, 3) and (5, 6, 7). 6. (-2, 3, -1) and (-3, -4, 3). 118. Angle between Two Lines. Let AB and CD, Fig. 79, be two lines, and let their direction cosines be cos aj, cos /3i, cos 7i, and cos 0:2, cos ^2, cos 72, respectively. Denote the angle between the lines by d. Let CH, HK, and KD be the edges of the parallelopiped formed by passing planes through C and D parallel to the coordinate planes. The projection of CD on AB is clearly equal to the sum of the projections of CH. HK, and KD on AB. Hence CD cos d = CH cos ai + HK cos /3i + KD cos 71. 232 CALCULUS [§119 Now and CH = CD cos a2, HK = CD cos /32, KD = CD cos 72. Consequently CD cos = CZ)(co8 ai cos ccz + cos /3i cos /32 + cos 71 cos 72). Hence cos 6 = cos ai cos 0:2 + cos |8i cos 182 + cos 71 cos 72. (1) Exercises Find the cosine of the angle between the lines determined by the points of Exercises 1 and 2; 2 and 3; 3 and 4, of the preceding section. 119. The Normal Form of the Equation of a Plane. — ^Let ABC, Fig. 80, be a plane. Let ON, the normal from 0, meet it in N. Let the length of ON be p and let its direction angles be a, j3, and 7. If p, a, /3, and 7 are given the plane is determined. We seek to find the equation of the plane. Let P, with co- ordinates X, y, and z, be any point in the plane. The sum of the projections of OH = x, HK = y, KP = z, and PN upon ON is ON = p. The projection of OH on ON is x cos a. The projection of HK on ON is y cos j8. The projection of KP on ON is z cos 7. The projection of PN on ON is 0. Hence . X cos a + y cos ^ + z cos 7 = p. (1) If P does not lie in the plane ABC, the projection of PN on OiV is not zero, and the coordinates of P do not satisfy (1). Hence the locus of a point satisfying (1) is a plane. Equation (1) is the normal form of the equation of the plane, p is taken to be §120] SOLID GEOMETRY 233 positive. The algebraic signs of cos a, cos/3, and cosy are de- termined by the octant into which ON extends. Illustration 1. Find the equation of a plane for which p = 2, a = 60°, 13 = 45°. cos a = ^, cos /3 = — -^. V2 Then by (1), §117, cos- 7 = 1 — J — §• Hence cos 7 = ± i. The equation of the plane is 2 + V2 - 2 There are thus two planes satisfying the conditions of the problem, one forming with the coordinate planes a tetrahedron in the first octant, the other a tetrahedron in the fifth octant. Exercises 1. Find the equation of a plane if a = 60°, /3 = 135°, p = 2, and if the normal ON extends into the eighth octant. 2. If a = 120°, /3 = 60°, p = 5 and if the normal ON extends into the sixth octant. 120. The Equation Ax + By + Cz = D. The general equation of the first degree in x, y, and z is Ax + By ^ Cz = D, (1) where A, B, C, and D are real constants. D may be considered positive. For, if the constant term in the second member of an equation of the form (1) is not positive it can be made so by dividing through by —1. .(2) Divide (1) by VA^ + B^ A ^ , + C2 B and obtain D VA^ -\- B^ + C^ Va^ 1 + B^ C + C2^ ^VI ' + B- 2 + C2 \/A^ + B^ + C^' 234 CALCULUS [§121 The coefficient of x is either equal to or less than unity in numerical value. It can then be considered as the cosine of some angle, say a. Similarly the coefficient of y may be considered as the cosine of some angle /3, and that of z as the cosine of some angle 7. Further the sum of the squares of these coefficients is equal to L Hence a,^, and 7 are the direction angles of some line. Then (2) is in the form X cos a -^ y cos /3 + 2 cos 7 = p, (3) where and cos a, cos /3, and cos 7 are the coefficients of x, y, and z, respect- ively, in equation (2). Hence (3) is the normal form of the equa- tion of a plane. Equation (1) is the general equation of the first degree in the variables x, y, and z. Therefore every equation of the first degree in x, y, and z represents a plane. Illustration 1. Put 3x — 2y — z = Q in the normal form. Divide by VZmTbM^ = \/9 + 4 + 1 = \/U and obtain 3j 2y z__ _ 6 Vii \/l4 Vii ~ Vii The plane is . — units distant from the origin, and forms, with the Vl4 coordinate planes, a tetrahedron in the eighth octant. Exercises Transform each of the following equations to the normal form, find the distance of each plane from the origin, and state in which octant it forms a tetrahedron with the coordinate planes. 1. Sx - 2y - z = 1. 6. X + 2?/ = 6. 2. X + y + z = —1. 7. x — 2 = 4. 3. X - 3r/ + 2z = 3. 8. x = 2. 4. x - 2j/ + 3z + 2 = 0. 9. X = -1. 6. 2x - y - z - 1 = 0. 10. z = y. 121. Intercept Form of the Equation of a Plane. We seek the equation of a plane whose intercepts on the X-, Y-. and Z-axes are a, b, and c, respectively. §121] SOLID GEOMETRY 235 The general equation of a plane is Ax + By -{- Cz = D. (1) The constants are to be so determined that the plane will pass through the points (a, 0, 0), (0, b, 0) and (0, 0, c). On substituting the coordinates (a, 0, 0), in (1), we obtain Aa = D, or a Similarly, since (1) passes through (0, b, 0), Bb = D, or -?• And, since it passes through (0, 0, c), Cc = D, or C With these values oi A, B, and C, (1) becomes a c or a ' b ' c + f + - = 1- (2) Equation (2) is known as the intercept form of the equation of a plane. Illustration. Transform the equation 3x — 2?/ — 52 = 4 to the intercept form. Divide by 4 and obtain -+-^ + -^=1 3 ■^ 5 The intercepts on the X-, F-, and Z-axes are i, —2, and — |, respectively. 236 CALCULUS [§123 Exercises Transform each of the following equations to the intercept form : 1. X +y +z = 3. 4. 2a; + 7y - 3z = 1 2. 2x - 3y + 42! = 7. 5. x - y + Sz = -\. 3. 2x + y - z + 2 = 0. 6. t/ - 2x - 3z = 5. 122. The Angle between Two Planes. The angle between two planes is the angle between the normals drawn to them from the origin. The cosine of the angle between the normals can be found by formula (1) §118, in which ai, /3i, 71 and 0:2, ^2, 72 are the direc- tion angles of the normals. Illustration. Find the angle between the planes x + y + z = l (1) and 2x + y + 2z = S. (2) Transform these equations to the normal form and obtain Vs Vs V^ Vs and 2a; V 22 3+1+3=1. (4) The direction cosines of the normals to the first and second planes are —t=, —j=, — ^, and f , \, f , respectively. Then, if Q is the angle between the normals, formula (1), §118, gives fl 2,1,2 5 cos Q = - — 7= + - — 7^ + 3\/3 3\/3 3\/3 3\/3 From which Q = 74.5°. • Exercises Find the angle between the following pairs of planes : 1. X - 3j/ + 2z = 6 and x - 2?/ + 2 = 1. 2. X - 2?/ + 3z = 2 and 2x + y - 2z = 3. 123. Parallel and Perpendicular Planes. If two planes are parallel = and cos Q — \. If they are perpendicular Q = 90° and cos 6 = 0. Let Arx-^ B,y + C,z = Di (5) and AiX + B^y + C2Z = D2 (6) §124] SOLID GEOMETRY 237 be the equations of two planes. After writing these equations in the normal form it is found that AiA2 + BiB2-\-CiC2 cos d = , (7) If AxA^ + B,B^ + C1C2 = 0, (8) cos B = Q and the planes (5) and (6) are perpendicular. If the planes (5) and (6) are parallel, the corresponding coeffi- cients must be equal or proportional. For then and only then will their normals be parallel. Exercises From the following equations pick out pairs of parallel planes and pairs of perpendicular planes. 1. a; + J/ + z = 6. 2. X — y - z = 2. 3. 2x + 2y + 2z = 7. 4. 3x - 27/ - z = 8. 6. 2x - 3i/ + z = 1. 124. The Distance of a Point from a Plane. Let {xi, iji, Zi) be any point and let Ax + By + Cz = D be the equation of a plane. We shall find the distance of the point from the plane. Now Ax + By + Cz = K, where K is any constant, is the equation of a plane parallel to the given plane. (See §123.) Let us choose K so that this plane shall pass through the given point (xi, ?/i, Zi). To do this substitute the coordinates of the point in the equation and solve for K. This gives K = Axi + Byi + Czi. Placing the equation of each plane in the normal form we have Ax-hBy + Cz D and R Ax + By + Cz K R R ~ R Axi + Byi + Czi R where R = VA^ + B^ + C. 238 CALCULUS [§125 The given plane is -h units distant from the origin, and the plane through the point (xi, yi, Zi) is p — units distant from the origin. Then the distance, d, between the two planes, and hence the distance of the given point from the given plane, is equal to the difference of these two distances, or d = Axi + Byi + Czi - D \/A24^B2"+C2 Illustration. Find the distance of the point (1, 2, —1) from the plane 3x — y + z-{-7 = 0. Axi + Byi -\-Czi-D _ 31 - 1-2 + !•( - 1) +7 ^ 7 Vn d = VA^ + B^ + C^ V32+(-l)2 + P Exercises In each of the following find the distance of the given point from the given plane: 1. (3,1,-2); 3x + y - 2z - 6 = 0. 2. (-1,2,-3); x-y -2z + l =0. 3. (0, 2,-3); 2x + 3y - 5z - 10 = 0. 125. Symmetrical Form of the Equations of a Line. Let PPi, Fig. 81, be a line passing through the given point Pi (xi, yi, zi), and having the direction cosines cos a, cos /3, cos y. In order to find the equations of the line, let P (x, y, z), be any point on the line and denote the distance PPi by d. Then Fig. 81. and therefore Xi X — Xi = d cos a, y - 2/1 = d cos /3, Z — Zi = d cos 7, y - 2/1 Z - Zi cos j3 cos 7 (1) COS a — These equations are known as the symmetric equations of the straight line. §125] SOLID GEOMETRY 239 Frequently a straight line is represented by the equations of two planes of which it is the intersection. Illustration 1. 3x-y + l=0, (2) 5x -z = S. (3) From these equations the symmetrical form of the equations can readily be obtained. From (2) and (3) we obtain _y-l _z+3 "^ ~ 3 - ^5~' or x-0 y-1 z + 3 ... nr = ~3~ = ~5- (^> The denominators, 1, 3, and 5, of (4) are not the direction cosines of the line, but they are proportional to them. Upon dividing each by \/35, the square root of the sum of their squares, they become the direction cosines. Then X -0 _ y-1 _ 2 + 3 1 ~ 3 ~ 5 -s/35 \/35 VS5 is the symmetrical form of the equations of the line. The line therefore passes through the point (0, 1,-3) and has the direction cosines given by the denominators in the preceding equations. Illustration 2. Consider the line which is the intersection of the planes 13a: + 52/ - 42 = 40, -13x + lOy -2z = 23. On eliminating x we obtain 5y - 2z = 21, and on eliminating y we obtain 13x -2z = 19. From the last two equations we find 5y - 21 13x - 19 ' = —2~ = 2 ' or x-n y-\^ z - A i 1 240 CALCULUS [§126 These are the equations of a line which passes through the point (+3, ^6*", 0) and whose direction cosines are proportional to A, 5, and 1. The student will find the direction cosines. In Illustration 1, equation (2) represents a plane parallel to the ^-axis whose trace in the XF-plane is the line 3x — 7j -{- 1 = 0. Equation (3) represents a plane parallel to the F-axis whose trace in the ZX-plane is the line 5x — z = 3. In Illustration 2 the position of the two planes which intersect in the straight line is not so evident. By eliminating first x and then y, the equations of two planes passing through the same line are obtained, one of which is parallel to the X-axis and the other to the y-axis. Exercises Put the equations of the following lines in the symmetrical form : 1. X + 2y +3z =6, X — y — 2 = 1. 2. a; + y — z = 1, a; - 3y + 2z =6. Z. X - y + 2z = 0, X + 2y - 3z =0. 126. Surfaces of Revolution. Let 2/2 = 42 (1) Fig. 82. be the equation of a curve in the FZ-plane, Fig. 82, and let it be rotated about the Z-axis. The sur- face generated is a surface of revolu- tion. Any point D on the curve de- scribes a circle of radius CD, equal to the 7/-co6rdinate of the point D. During the revolution the z-co6rdinate does not change. Let P be any posi- tion taken by D in the revolution. Let the coordinates of P be (x, y, z). But by (1), x2 -f 7/2 = icpy = {coy {CDY = iz, (2) §126] SOLID GEOMETRY 241 where z is the common ^-coordinate of Z) and P. Then (2) becomes x2 + 2/2 = 42, (3) an equation satisfied by any point on the surface of revolution. We note that (3) is obtained from (1) by replacing ^/^ by x^ + y"^, or y by -s/x^ + y"^. In general, if /(y, 2) = (4) is the equation of a plane curve in the FZ-plane, the equation of the surface of revolution generated by revolving it about the Z- axis is obtained by writing -s/x^ + y"^ for y, i.e., the equation of the surface of revolution is fiVx^TV', ^) = 0- (5) This equation can also be regarded as the equation of the surface generated by revolving the curve f{x, z) = 0, lying in the XZ- plane, about the Z-axis. Similarly, /(i/, V^M^) = o (6) is the equation of the surface generated by revolving the plane curve /(y, x) = about the F-axis; and <A(x, V^M=^) = (7) is the equation of the surface generated by revolving the plane curve 0(.T, 2) = about the X-axis. Illustration 1. The equation of the surface generated by rotat- ing x^ -{• iy — ^y = a^ about the X-axis is x^ + [Vy^ + 2' - ^Y = a^ Exercises Find the equation of the surface generated by rotating: 1. y = x^ about the F-axis. 2. y =^ x^ — a^ about the X-axis. 3. 6^x2 + a^yi = a^b^ about the X-axis. 4. 6^x2 - 0^2/2 = a^b^ about the X-axis. 6. &2x2 — a-y^ = a-b^ about the F-axis. 6. x2 + y^ = a^ about the F-axis. 7. x2 -f- t/2 = a2 about the X-axis. 8. y = mx about the X-axis, 9. J/ = mx about the F-axis. 16 242 ■ CALCULUS [§127 127. Quadric Surfaces. Any equation of the second degree between x, y, and z, of which Ax^ + By^ + Cz^ + Dxy + Eyz + Fxz -\- Gx -{- Hy + Kz -h L = (1) is the general form, represents a surface which is called a quadric surface, or conicoid. By a suitable rotation and translation of the axes, the equation of any quadric surface can be put in one of the following forms: • x^ v^ z^ ± ^' ± P ± ^ = 1' (2) rp2 y2 2*2 <^ ± P ± C-' = »■ (3) X^ 7/2 -,±i,= + 2c2. (4) The particular form assumed by the equation depends upon the values of the coefficients in (1). The quadric surface X^ 0/2 2j2 ^= + f= + ,T = l (5) is called the ellipsoid. To find the shape and properties of this surface, let X = k, (6) where k is any real constant. This equation represents a plane perpendicular to the axis of x. Equations (5) and (6) considered as simultaneous equations represent the curve of intersection of the ellipsoid with the plane. If x is eliminated between (5) and (6) there results t/2 ^2 / by/a^-k^ y ^ / cVa^ - k^ V the equation of the curve of intersection in the plane x = k. Equation (7) is the equation of an ellipse. The semi-axes of the „. &\/a2 - k^ , c\/a^ - k^ rru u * ellipse are —^ and —^ . These axes grow shorter a, a as A; increases in numerical value from to a. When k = ± a §127] SOLID GEOMETRY 243 the elliptical section reduces to a point. When | A;| > a, the lengths of the axes of the ellipse become imaginary, i.e., the plane x = k, (,\k\ > a), does not meet the surface (5) in real points. Hence the surface is included between the planes z = + a. The above discussion shows that the surface represented by the equation (5) is included between the planes x = + a; is symmet- rical with respect to the FZ-plane; and has elliptical sections made by planes perpendicular to the axis of x. These sections grow smaller as the cutting plane is moved away from the YZ- plane and at a distance + a reduce to a point. In a similar manner, by taking y = k, and then by taking z = k, the student will discuss plane sections of the ellipsoid (5) perpendicular to the F-axis and to the Z-axis. a, b, and c, are called the semi-axes of the ellipsoid. It can be shown that any plane section of the ellipsoid is an ellipse. The surface represented by /2 ^^ . r _ 1 _ 1 rsi ^2 + ^,2 c^ - 1 - w will now be discussed. Let z = k. Then S + f: = i + l' (») is the equation of the plane section made hy z = k. It is an ir . ■ aVcM^2 , bVc^ + k^ rp, ellipse whose semi-axes are — ' and — ^^ ■ . They increase in length with the numerical value of k. The axes have a minimum length when k = 0. The surface represented by equation (8) is symmetrical with respect to the XF-plane, and every section parallel to this plane is an ellipse. The smallest elliptical section is that made by the XF-plane. li X = k, equation (8) becomes P - c"^ = 1 - a^' (10) an hyperbola. If A; < a, the transverse axis of the hyperbola is parallel to the 244 CALCULUS [§128 F-axis. If fc > o, the transverse axis is parallel to the Z-axis. When k = a, equation (10) reduces to or r z'^^^ (^+^)(i-:-)=». the equation of two straight lines. The student will discuss the curves of intersection of the surface (8) with planes parallel to the XZ-plane. The surface is called the hyperboloid of one sheet, or of one nappe. Exercises The student will discuss the following surfaces and make sketches of them: a;2 y2 z2 1. —2~ hi 2 ~ ^» ^^^ hyperboloid of two sheets. 2. ~^— Ti = 2cz, the hyperbolic paraboloid. -j;2 y2 3. — 5 + r^ = 2c2, the elliptic paraboloid. 128. Cylindrical Surfaces. If the circle x2 + 2/2*=25 (1) be moved parallel to itself so that all of its points describe lines parallel to the Z-axis, it will generate a right circular cylinder. The equation of this cylinder is sought. In any plane z = k, the relation between x and y for points in the curve of intersection of this plane and this cylinder is the same as that for points in the plane z = 0, viz., x"^ -{- y^ = 25. Now, this equation is satisfied by all points on the surface for all values of z. Hence it is the equation of the surface. The cylindrical surface just considered can be regarded as generated by a line moving parallel to the Z-axis and passing through points of the circle x^ -\- y^ = 25 in the plane z = 0. In general a cylindrical surface is a surface generated by a line moving parcdlel to itself. §128] SOLID GEOMETRY . 245 It is clear that the equation Kx, y) = (2) represents the cylindrical surface generated by a line moving parallel to the Z-axis and passing through points of the curve f{x, y) = in the plane z = 0. The equation of a section of (2) made by any plane z = A; is j{x, y) = 0. Thus $ + $-^ (3) represents an elliptical cylinder whose elements are parallel to the Z-axis. 3.2 _j_ 1^2 _ 2ax (4) represents a circular cylinder whose elements are parallel to the Z-axis. The center of the section in the plane z = is the point (a, 0). By the same reasoning 7/ + z^ = a^ (5) represents a circular cylinder whose elements are parallel to the X-axis. z2 = 4x (G) represents a parabolic cylinder whose elements are parallel to the y-axis. The plane a; - 4y + 3 = (7) can be regarded as a cylindrical surface whose elements are parallel to the Z-axis and which pass through the line X , 3 y-i + i in the plane z = 0. In general, an equation in which one of the letters x, y, z is absent, represents a cylindrical surface whose elements are parallel to the axis corresponding to the letter which does not appear in the equation. 246 CALCULUS I§129 Exercises Describe the surfaces represented by the following equations: 1. x^ + y^ = 16. 7. x« - 2/2 = 0. x^ V^ 2- 4 +16 = 1- 3. a;* - t/« = 1. 4. 2» + y2 = 25. 5. 22 - X* = 25. B. X + 2y = 10. 8. a;?/ = 1. 9. xz = 2. 10. (x -3)(x + 2) = 0. 11. y^ = 4x. 12. 7/2 + 2" = 2ay. 13. x2 + 2/2 = lOx. 129. Partial Derivatives. Let z = f(x, y) be a function of two independent variables, x and y. When x takes on an increment Ax, while y remains fixed, z takes on an increment which we shall denote by AxZ. When y takes on an increment, Ay, while x remains fixed, z takes on an in- crement which we shall denote by A^2. For example, if a gas be en- closed in a cylinder with a mov- able piston, the volume v of the gas is a function of the tempera- ture T and of the pressure p which can be varied by varying the pressure on the piston. If the temperature alone be changed the volume will take on a certain increment A?^. If the pressure alone be changed the volume will take on the increment ApV. If 3 = /(x, y) be represented by a surface. Fig. 83, the increment of z obtained by giving x an increment, while y remains constant, is the increment in z measured to the curve cut out by a plane y = k, Q, constant. Thus A^z = HQ; similarly A^z = KR. AxZ The limit of the quotient -r— as Ax approaches zero is called the partial derivative of z with respect to x. It is denoted by the dz symbol -t^. Then dx dz dx lim AxZ ^=0 Ax* §130] SOLID GEOMETRY 247 It is evidently calculated from z = J{x, y) by the ordinary rules of differentiation, y being treated as a constant. Thus if z = x^y, dz Geometrically w— represents the slope of the tangent at the point (x, y, z) to the curve cut from the surface by the plane through this point parallel to the XZ-plane. Similarly dz_ _ lim A^ dy ^J/=o Ay and it is calculated by differentiating z = /(x, y), treating x as a constant. Geometrically it represents the slope of the tangent at the point {x, y, z) to the curve cut from the surface by the plane through this point, parallel to the FZ-plane. If z = xhj, ■5— = x^. dy X Illustration 1. If 2 = sin -> y dz X d /x\ 1 X = cos - -^T" I - ) = - cos - dx y dx \yj y y and dz X d /x\ X X V- = cos - ^- ( - ) = ~ cos -• dy y dy \y} y^ y 130. Partial Derivatives of Higher Order. If z is differentiated twice with respect to x, y being treated as a constant, the deriva- tive obtained is called the second partial derivative of z with re- dh spect to x. It is denoted by the symbol 3—^. Similarly the second partial derivative of z with respect to y is denoted by the symbol dy^' If z is differentiated first with respect to x, y being treated as a constant, and then with respect to y, x being treated as a constant, dh the result is denoted by the symbol „ „ • If the differentiation takes place in the reverse order the result is denoted by the symbol 248 CALCULUS [§130 - „ • The first is read "the second partial derivative of z with dxatj respect to x and y;" the second, "the second partial derivative of z with respect to y and x." In the case of functions usually occurring in Physics and Engineering, viz., functions which are continuous and which have continuous first and second partial d-z d'^z derivatives, „ ^ ■ = ^ „ • The order of differentiation is dyax oxay immaterial. Illustration 1. z = xhj. dz ^ d^z „ dh ai = 2^^^' d^^ = 2^' d^x = 2^- ^ ^ 3.2 ^ = -^- = 2x ^2/ ' dy^ ' dxdy In this case d^z _ d^z dydx dxdy X IlliLstration 2. z = dz dx " = sin • y 1 X = - cos — y y dx^ " 1 . X = 2 sin • y' y a?/ax 1/ . x\ / x\ 1 X = - ( — sin I ( — „ ) — „ cos ~ y\ yJ \ yV 2/' 2/ = 1 / . X x\ = , I a; sin w cos - ) • y'\ y ^ y) dz dy ' X X = i cos -• 2/' y d^z dy'' 2x X x' . X = 7 cos iSin -• 2/^. 2/ 2/^ 2/ . dH dxdy a; / . a;\ /1\ 1 x = —i. I sm " I I - I i cos 2/^ \ y) \yJ 2/' y = 1 / . X x\ = —. I a; sin y cos - ) • 2/' \ y y) Here again, we notice that d'z dH dydx ~~ dxdy §130] SOLID GEOMETRY 249 Exercises d^z d^z d'^z d^z 1. Find ^r^» ^r— ,' -> a ' and ^ a ' for each of the functions : (a) z = ^- (b) z = xy\ (c) z = xhj. (d) z = sin xy. (e) z = e* sin y. 9^z 6^z 2. Find , - - and - ^ for each of the following functions : dydx dxdy (a) z = x^y. (h) z = xsin'^y. (c) z = x cos y. Id) z = y log X. (e) z =- e" sin x. (J) z = y tan'i j;. It is seen that ^-^- = ^-^T in all of these cases. dydx dxdy In the above discussion z was considered to be a function of two independent variables only. The notion of partial derivatives can, however, be extended to functions of three or more variables. Illustration 3. li z = xhjt, ix = 2^^^' dy "" '' dt = ^ ^' and dy _^ dxdt " *'^^' d'z ^2x ' dtdydx CHAPTER XV SUCCESSIVE INTEGRATION. CENTER OF GRAVITY. MOMENT OF INERTIA 131. Introduction. In the preceding chapters there have been numerous examples of successive integration of functions of a single independent variable. Thus, to determine the law of motion of a falling body whose differential equation of motion is it is necessary to integrate twice. The result of the first integration ds is -^ = gt -{- Ci, and that of the second is s = ^gt^ + Cit + Cz. Exercises 1. If -j—^ = 2x, find y as a function of x, given that -r- = 3 when X = 1, and y = 2 when x = 4; given that ?/ = 4 when x = 2, and that y = 7 when x = 4. d^y 2. Find y if -7-^ = 7x. Assign suitable conditions to determine the constants of integration. 3. Find y if ^ = 2x\ The operation of finding the result of Exercise 2 can be written J[ JiJtx dx\dx]dz = /[/[la;^ + Ci\dx\dx = ^ + Ci^- + C,x + C3. The first member can be written Ix dx dx dx. ///' It is a triple integral and indicates that integration is to be per- formed three times in succession. An arbitrary constant of 250 §132] SUCCESSIVE INTEGRATION 251 integration is introduced with each integration. If each integra- tion is performed between limits the constants of integration do not appear. Thus, 7x dx dx dx = I I -^ dx dx is: 42 dx dx = I 42 a; dx ■r 1 = 1 84 dx |2 = 84 X =84. |i 132. Illustration of Double Integration. Let il = -' + y- <i) Integration with respect to y, treating x as a constant, gives ^ = x^t/ + | + 0(x), where (t>(x) is an arbitrary function of x. This arbitrary function of X takes the place of an arbitrary constant of integration in the case of a single independent variable. A second integration, this time with respect to x, gives 2 = ^ + ^' + J<A(^)rfx + ^(^), (2) where 4'{y) is an arbitrary function of y. The result contains an arbitrary function of x and an arbitrary function of y. Equation (2) represents a surface, but a very arbitrary one on account of the presence of the arbitrary functions I (f){x)dx and 4'(.y)- The process of finding (2) from (1) is indi- cated by a double integral sign. Thus, // (x2 + ?/2) dy dx, 252 CALCULUS [§132 which means I ( I (x2 + 2/2)di/) dx. Upon performing the integration indicated, first with respect to y, then with respect to x, we obtain I I (x2 + t/2) dydx = I (x^y +J + <f>ix)) dx Instead of an indefinite double integral such as the one just considered we may have a definite double integral. If the inte- gration with respect to y is performed before that with respect to X, the limits of integration with respect to y may be functions of X. Thus, I \\x^ + y') dydx=f (xH) + I) Y dx The last integration is readily j)erformed. It is to be noted that in evaluating a double integral, x is treated as a constant when the integration with respect to y is performed. If in a definite integral dx is written before dy, the integration with respect to x is to be performed first. ^ 1. 1 1 xydydx. Exercises f*ir f*a{\ + C03 B) 4. 1 1 rdrde. Jo Jo 2. 1 1 xy dy dx. Jo Jo r- rva-^-x^. 5. 1 dydx Jo Ja—x 3. 1 1 xy-'dydx. Jo Jix 6. 1 1 1 x^y^z^ dz dy dx. Jo Jo Jz 1 Usage varies on this point. The student will have to observe in every case the convention adopted in the book he is reading. §133] SUCCESSIVE INTEGRATION 253 I dzdy Jo nVx n rvi xdydx. 10. I I n'' {x^ + y^) dy dx. 11- I I 2 Jo Jx^ dx. 10. 12. dy dx. y dy dx. Vo^ — ^"^ — y^ dy dx. Hint. To perform the integration in Exercise 12, let v a^ — x^ = b and make use of the result of Illustration 3, §105. 133. Area by Double Integration: Rectangular Coordinates. A plane area can be represented by a double integral. Thus, let it be required to find the area A between the curves y = /i(x), y = f^ix), and the lines x = a and x = b. The area of the strip IJKH, Fig. 84, is approximately lim Ay «2 «t /.V, j2)At/Aa; = Aa;]^m^2)Ay = Ax I dy, where yi and y^ are the ordinates of the two curves y = /i(x) and y = fiix), respectively. And the area sought is approxi- mately x = a Jvi Fig. 84. The smaller Ax is taken, the "o closer the approximation. The limit of this sum as Ax approaches zero is the area sought. Since y\ and 2/2 are func- tions of X, 1 dy is, a, function of x and consequently lim K)X Ax I ^2/= I \ dy dx = A. It is to be noted that in setting up this integral the summation 254 CALCULUS [§133 with respect to y was performed first, giving the area of a vertical strip for a particular value of x. Consequently, the integration with respect to y is to be performed first, x being treated as a constant. On performing the integration with respect to y we obtain (2/2 - yi)dx = I [/zCx) - Si{x)\dx, a Ja a single integral which might have been set up at once by consider- ing the area as the sum of vertical strips of length 1/2 — yi, and of width Ax. It is, however, desirable to be able to set up a double integral over an area. In choosing the limits for a double integral, the student should proceed systematically. The process of setting up the above integral with its limits is the follow- ing: The "element" is the rectan- gular element of area dy dx. The "summation" (integration) of this element, for a particular value of x, between the limits for y of WI and WH, the ordinates of the curves y = /i(x) and y = /2(.t), gives the area of the typical strip IJKH. The "sum- mation" (integration) of the strips of which this is a typical one, between the extreme values oi x, x = a and x = b, gives the area sought. Thus A = I I dy dx. nv The procedure may be briefly summarized in the following concise directions. Write first the element dy dx, then the integral sign, then the limits /i(x), /2(x), then another integral sign with the limits a and b. Illustration. Find by double integration the area between the parabolas y^ = x and y = x^. The integral is set up as follows: Write the element dy dx, then an integral sign with the limits x^ and \/x. This represents the area of the typical strip, IJKH, Fig. 85, for a fixed x. All of the strips of which this is a typical §134] SUCCESSIVE INTEGRATION 255 one are to be summed from a; = to x = 1, the abscissas of the points of intersection of the curves. Then write the second integral sign preceding the first with the limits and 1. Thus Jo Jx' dy dx = \. Exercises 1. Find by double integration the area between the curves y = x and y^ = x'. 2. Find the area of Exercise 1 by integrating first with respect to x and then with respect to y. 3. Find by double integration the area between r/* = a{a — x) and X* + 2/^ = a^. 4. Find the area between y^ = ax and y^ = 2ax — x^. 5. Find the area of Exercises 3 and 4 by integrating first with respect to X. 6. Find the area bounded by y^ = 4x, x + y = 3, and the X-axis. 7. Find the area of a rectangle by double integration. 8. Find the smaller area between x^ + y^ = 1 and y = x + 5- 134. Geometrical Meaning of the Definite Double Integral Consider the definite double integral fix, y) dy dx. (1) , = /i(x) J'»6 /»2/2=/20 In accordance with the definition of a definite single integral, §66, (1) can be written f'[ll^oXf(^'y) ^y]d^' (2) Here x is considered constant under the summation sign, and f{x, y) is, for such a fixed x, a function of y alone. V, lim ^2 /»2/2 is a function of x, since x occurs as an argument of / and also in the limits of integration. Hence we can write (2) in the form 256 CALCULUS [§134 lim O L «. J -E%'Zll%Xn-.y)^vAx, (3) where Ax under the second summation sign is regarded as a con- stant multiplier. Z Fig. 86. In Fig. 86, let EFGL represent the surface z = f{x, y) ; QABS, in the ZF-plane, the curve y = fi{x) ; DHKC, the curve y = fi{x) ; AD the line x = a; BC the line x = b; and A'B'C'D' the portion of the surface cut from z = f(x, y) by the cylinders y = fi{x), y = f^ix), and the planes x = a and x = b. §134] SUCCESSIVE INTEGRATION 257 Divide A BCD into small rectangles, as shown in the figure, by lines parallel to the X- and F-axes, at intervals of At/ and Ax, respectively. Through these lines pass planes parallel to the XZ- and FZ-planes. These planes divide the solid bounded by the planes and surfaces of Fig. 86 into vertical columns of rectangular cross section Ay Ax. The column erected on MNPR as a base is a typical one. f{x, y)Ay Ax represents approximately the volume of the column whose base is MNPR and whose top is M'N'P'R', since the area of its base is Ay Ax and its altitude is MM' = f{x, y). Then the sum of the columns at a fixed distance from the yZ-plane, 2)/(x, y) Ay Ax, is approximately the volume of the slab between the planes IHH'I' and JKK'J', i.e., between the planes x = x and X = X + A X. And h r «2 -1 a L w. -I the sum of the volumes of all the slabs, is approximately the vol- ume of the solid ABCDA'B'C'D'. If Ay and Ax are each taken smaller and smaller this sum will eventually represent a very close approximation to the volume in question, and the limit of this sum as Ay and Ax approach zero is the volume. Hence the inte- gral f(x, y) dy dx, which we have seen is equal to X = b r- V a; = a L v. J lim x = a represents the volume bounded by the plane z = 0, the surface ^ = /(-^j y), the planes x = a and x = h, and the cylinders y = /i(.t) and y = Jiix). Illustration. Find the volume contained in the first octant of 17 258 CALCULUS [§135 the sphere x^ + y^ + z"^ =- a^. See Fig. 87. The equation of the surface is z = \/a2 — x^ — y^. yi = fiix) = 2/2 = f2{x) = Va^ - x\ the trace of the sphere on the XF-plane. The volume of the col- umn on MNPR as a base is Va^ — x^ — 2/2 A?/ Ax, or, as we shall say in the future, \^a^ — x^ — y^ dy dx. The summation of these columns for a fixed x gives Fig. 87. r ■y/a'^ — x^ — y"^ dy dx, !/■ the volume, expressed as a function of x, of the slab between the planes x = x and x = x4-Axorx = x + dx. The summation of all these slabs from x = to x = a gives *y/a^ — xi ^/a^ — x^ — y^ dy dx, to the volume of one octant of the sphere. This integral was evalu- ated in Exercise 12, §132. Exercises 1. Find the volume of the segment of the paraboloid y^ + 2z^ = 4x, cut off by the plane x = 5. 2. Find the volume bounded by the cylinders y = x^ and y- = x, and the planes z = and z = 1. 3. Find the volume common to the cylinders x^ + y^ = a^ and y^ + z^ = a2. 4. Find the volume between the cylindrical surface y^ — x', the plane y = x, and the planes z = and z = 1. 135. Area: Polar Coordinates. Let it be required to find, by double integration, the area between the radii vectores 6 = a, §135] SUCCESSIVE INTEGRATION 259 and d = ^, and the curve p = f{d). Divide the area as shown in Fig. 88, the radii making an angle of Ad with each other and the radii of the concentric circles differing by Ap. The area of MPQR is equal to Kp + Ap)2 A0 - § p2 A^ = p Ap A0 + ^ Ap2 A0 As Ap approaches zero, limpApA0 + |Ap2A5 Hence P=Ke) ^"=0 p Ap A^ = 1. p = p = Jp = Fig. 88. Fig. 89. This sum represents the area of the sector OHL. The total area sought is the limit of the sum of these sectors as A9 approaches zero, i.e. ^ = Ai"oX^^ p^p= \ p^p^- This integral is to be set up as follows: The element of area is the approximately rectangular area MPQR whose area is approxi- mately {MR){MP) = pdpdd. This element is to be summed from p = to p = f{d) . This gives approximately the area of the typical sector OHK. These sectors are to be summed from e = atod = fi. 260 CALCULUS [§136 More briefly: Write down the element pdpdd, then an integral sign. Its limits are the extreme values of p for a given 6. Then write before this integral another integral sign. Its limits are to be chosen so as to sum up all the sectors such as OHK. Illustration 1. Find the area of the circle p = 10 cos 6, Fig. 89. The area bounded by the semicircle above the initial line will be found and multiplied by two. .20 cos ff Jo Jo lOcosO A =2\ I pdpde. p dd dp. Illiistration 2. Find by double integration the area between p = 10 cos e and p = 20 cos d. See Fig. 90. •20 COS e A =2\ I pdpdd 'l0CO3 tf Show that t/O t/10 mo /•«=o«-'^ /•20 /-cos-^ A = 2 I I ^ pdddp+2 \ I pdddp. Jo JC08-1-^ JlO Jo 10 Which method is the simpler in this case? Exercises Find by double integration: 1. The area of the circle p = 5 sin 0. 2. The area of the cardioid p = a(l — cos 9). 3. The area of the lemniscate p" = a" cos 20. 4. The area outside p = a (1 + cos 6) and inside p = 3a cos d. 136. Volume of a Solid: Triple Integration. We shall now find the volume of the solid of Fig. 86 by triple integration. Sup- §136] SUCCESSIVE INTEGRATION 261 pose the solid further subdivided by planes parallel to the XY- plane and at a distance Az apart, into rectangular parallelopipeds of Volume Az Ay Ax. Then the volume of the column on the base MNPR is approximately a = f(.x, y) ^ Az Ay Ax. Z = Then Vi z => f(.x,y) '^ 2^ Az Ay Ax y\ « = is approximately the volume of the slab between the planes IHI'H' and JKJ'K', i.e., between the planes x = x and X = X ■\- Ax. X = h 2/2 2 = f{x, y) X 2) 2) Az Ay Ax X = a y^ « = is approximately the sum of the volumes of all of the slabs. If Az, Ay, and Ax are each taken smaller and smaller, this sum will represent a very close approximation to the volume sought, and the limit of this sum as Ao:, Ay and Az approach zero is exactly this volume. Hence the integral. a Jy^ Jo dz dy dx, represents the volume bounded by the plane z = 0, the surface z —j{x,y), the planes x = a and x = h, and the cylinders y = j\{x) and y = Si{x). Illustration 1. Find by triple integration the volume of the ellipsoid a" ^ 62 -r g2 See Fig. 91. F = 8 I I dzdydx Jo Jo Jo 262 CALCULUS [§136 The student will perform the integration. Illustration 2. Find by triple integration the volume of the solid bounded by the cylinder x^ -\- y^ = 2ax, the plane z = 0, and the paraboloid of revolution x^ -{- y^ = ^az. Write the ele- ment of volume, dz dy dx. The integration with respect to z cc^ "4* 1/^ -J gives the volume of the typical between the limits and vertical column of base dy dx, extending from the point (x, y) in /j;2 _|_ y2 the plane z = to the surface of the paraboloid, z = — j ' Next, X being kept fixed, these columns are summed into a typical slab by integrating with respect to y from the Z-axis, y = 0, to y = ^/2ax — x^, the trace of the cylinder in the XY- plane. Finally the integration with respect to x from x = to X = 2a gives one-half of the total volume sought, viz., that lying in the first octant. xi + y^ F = 2 I I I dzdy dx. Jo Jo Jo The student will perform the integration. Exercises 1. Find the volume common to the cylinders x" + y^ = r^ and x2 -f- 22 = r^ §137] SUCCESSIVE INTEGRATION 263 2. Find the volume of one of the wedges cut from the cylinder ^2 ^ y2 — J.2 \yy ^hc plancs z = and z = mx. 3. Find the volume in the first octant bounded by the coordinate X y z planes and the plane — h r H — = 1. ^ ^ a c 4. Set up the integral representing the volume bounded by the surface x^ -\- y^ + z^ = o*. 6. Find the volume between y^ + z^ = 4ax and x — z = a. 6. Find the volume between the planes y = 0, z = and the sur- faces z = x^ + Ay^, y = 1 — x'^. 7. Find the volume between y^ + 2z^ = 4x and z = x. 137. Center of Mass, Centroid. Let there be a system of masses mi, m2, ms, . . . , rUn situated at the points {xi, yi, Zi), {xi, 2/2, 22), (^3, 2/3, Zz), . . . , {Xn, yn, Zn) , Tespectlvely. The mean distance with respect to mass, of the system from the FZ-pIane is The mean distances, with respect to mass, of the system from the ZX- and XF-planes are, respectively, 2 m.i/< and (2) (3) The point (x, y, z) is called the centroid, or the center of mass, of the system of masses mi, m2, • • • , m„. m,- Xi is called the moment, and Xi the moment arm, of the mass mi with respect to the FZ-plane.^ Then x is the mean moment arm with respect to the FZ-plane of the masses mi, m2, • • •, m„. For, equation (1) shows that if all the masses were placed at the distance, x, from the FZ-plane, the moment with respect to this plane would be the same as the sum of the moments of the masses. Hence we can say that the centroid of a system of masses is a point such that if all the masses were concentrated at this point, the moment with respect to each coordinate plane would be equal to 1 The term moment of a mass with respect to a plane has evidently a diflerent significance from the term moment as applied to a force. 264 CALCULUS [§139 the sum of the moments with respect to the corresponding planes of the masses in their given positions. 138. Centroid Independent of the Position of the Coordinate Planes. It will now be shown that the distance of the centroid, (x, y, z), from any plane is the mean of the distances, with respect to mass, of the masses mi, m2, • • •, m„, from that plane. And thus it will be shown that (x, y, z), the centroid, is a point whose position with reference to the masses is independent of the choice of the coordinate planes. Let ax -\- hy -\- cz -\- d = hQ the equation of a given plane (see §120). The distance, p, of the point (x, y, z) from this plane is - ax + hy + cz ■\- d P = 22 ' ^^^ where R = Va^ + 62 + c\ (See §124.) On substituting the values of x, y, and 2 from (1), (2), and (3) §137, and reducing the absolute term, d, to the common denominator, we have aSwiX,- + blfTTiiyi + cZniiZi + dXnii or sm.r R'Emi axi + hyi + czi + d' R J . (2) ox,- + hyi •\-czi + d • +1 j- . f .1 • ^ But 5 = pi is the distance of the point K (x,-, yi, Zi) from the given plane. Hence (2) can be written in the form _ S mipi This proves the statement at the beginning of this section. In other words, if all the masses of the system were concentrated at the centroid, the moment with respect to any plane would be equal to the moment of the system of masses with respect to this plane. 139. Center of Gravity. Let the system of masses considered above be acted upon by gravity. It will be shown that the line §140] SUCCESSIVE INTEGRATION 265 of action of the resultant force passes through the center of mass, or the centroid. Since the position of the centroid is independent of the choice of axes, choose the positive direction of the axis of z vertically upward and the axes of x and y horizontal. The force acting on mi is mi^, that on m^ is m-ig, etc. The resultant force is equal to 2 niig and is directed vertically downward. Its line of action meets the XF-plane in a point (a, j3, 0) such that its moment, aSm.gr, about the F-axis is equal to 2m,gfXi, the sum of the moments of the forces acting on the individual masses; and such that the moment, jSSmigr, about the X-axis, is equal to the sum of the moments, Sw.grr/,, of the forces about this axis. and Whence and Consequently Hence the resultant passes through the centroid of the system of masses. If the masses all lie in one plane, say the XF-plane, z is zero and the centroid is fixed by the two coordinates z and y. The product rtiiXi is called the moment of the mass rrii with respect to the F-axis. In this case x is the mean moment arm with respect to the F-axis. If the masses all lie upon a line, say the X-axis, the centroid is fixed by a single coordinate, x. 140. Centroid of a Continuous Mass. If instead of discrete masses we have a continuous mass, the coordinates of the center of mass, or the centroid, are clearly. a2,m iQ = 'Lrrii g^i, /SSw xQ = 'Lrui gvi- ^rUiXi Oi = ^rrii ^ = ^rmyi 2)^< a = X and |8 = y- 266 CALCULUS [§140 Ani = X = y = i%^Am Jdm z = IL^.oX^ /dm ^jj Vz Am J z dm ^ X -^m fdm lim Am Am=( The integration is to be extended throughout the entire mass, and the integrals considered may be single, double, or triple, depending on the form of the mass. Illustration 1. Find the center of gravity of a bar, Fig. 92, of length L, whose linear density, p, may vary. Let the axis of x coincide with the bar, the origin being taken at one end. The OF ^1 Fig. 92. mass of an "element" of the bar of length dx is p dx, p being a function of x, the distance of the element from the origin. The moment of this element of mass, dm = p dx, about an axis through the origin perpendicular to the bar is X dm = xp dx. X, the abscissa of the centroid, the only coordinate necessary to fix the centroid in this case, is given by px dx mJ X = f. p dx The numerator represents the total moment, and the denominator the total mass. If the bar is of uniform density, p can be taken il40] SUCCESSIVE INTEGRATION 267 out from under the integral sign. Then xdx — 2 X = r ^L. dx x\ If the linear density is proportional to the distance from one end, then p = kx and we have k \ x^ dx 1 f k f. X dx Illustration 2. Let it be required to find the center of gravity of a plate of uniform thickness and of mass p per unit volume or of mass p per unit surface. Take a plate of the shape of Fig. 84. The mass of the element MNPR is p dy dx. The moment of this element about the Y- axis is xp dy dx, and its moment about the X-axis is jj p dy dx. Then r r p X dy dx fdm r p p^y^^ p y dy dx If p is constant, and i ^^« r r^pdydx Ja Jvi I xdy dx I { dy dx Ja Jy^ J'*b nvi I ydydx a Jy^ y^ rb /»!/, ' II dy dx Ja Jv^ 268 CALCULUS [§140 The numerator of each of these expressions is the integral of the product of an element of area by its distance, x or y, from the Y- axis or X-axis, respectively. The denominator is the area. The mass does not enter into either of these formulas. We are thus led to speak of the centroid of an area, of a line or of a solid, with- out reference to its mass. This notion of the centroid of a geo- metrical figure, a line, an area, or a solid, without reference to its material composition is an important one. For, in many prob- lems in mechanics one is interested in the centroid of a geomet- rical configuration as such. Thus in the study of the deflection of beams it is necessary to know the position of the centroid of the cross section of the beam. Illustration 3. Find the centroid of the solid represented in Fig. 86. The element of mass, dm, is equal to p dz dy dx, and its moment with respect to the yZ-plane is xp dz dy dx. Then Vix, y) px dz dy dx I X dm I dm Ja Jy^ Jo Similarly : Ja Jy^ Jo J Jo 'fix, y) p dz dy dx '/(x, y) py dz dy dx and _ \y dm J^^ I Pdzdydx Ja Jvi Jo j zdm 1 dm rb ny, n Ja Jvi Jo 'Kx, v) pz dz dy dx rb ry, r Ja Jy^ Jo iix, y) p dz dy dx If the density is constant, p can be canceled from numerator and denominator. If the solid has an axis or a plane of symmetry the centroid lies in this axis or in this plane. Illustration 4. Find the centroid of the area in the first quad- rant bounded by the circle x^ -\- y^ = a^. §140] SUCCESSIVE INTEGRATION 269 If we use double integration we have, in accordance with Illus- tration 2, X dy dx n — Jo Jo 4 and ydydx y = — "—^ 4 Radicals could be avoided in the evaluation of the numerator of the expression for x if the integration were performed first with respect to x and then with respect to y. Thus "Va2 - j/2 X dx dy '0 «/o X = ^ • 1" The student will evaluate each expression given for x. From the symmetry of the figure, x = y, and it is not necessary to evaluate the integral for y. In finding the centroid in this case, and indeed in many cases, it is easier to use single integration than double integration. Thus if we choose as the element of area, the strip y dx parallel to the y-axis, the moment of this strip about the F-axis is xy dx, and X dm I xy dx I x\/a' — x^ dx Jo Jo h f ydx -^ Illustration 5. Find the centroid of the solid in the first octant bounded by the sphere x^ -\- y^ -\- z^ = a^. The method of Illustration 3 gives I X dz dy dx Jo Jo Jo ' X = -, 270 CALCULUS (§140 From considerations of symmetry, y = z = x. Here again it is simpler to use single integration. Choose as element a slab of thickness dx parallel to the FZ-plane. The base of such a slab is a quadrant of a circle of radius \/a^ — x^, where x is the distance of the slab from the FZ-plane. The volume of this elementary slab is 7r(a^ — x^) , z ax. Hence TT X = I X {a^ . , .^ . V. x^) dx 4. f ^ Exercises Find the coordinates of the centroid of: 1. The area between y = x^ and y^ = x, 2. The areas of Exercises 1, 3, and 6, §133. 3. A triangular plate. Hint. Draw lines parallel to the base, BC, Fig. 93, at intervals dx along the median AM. The mass of each strip is proportional to AL = X and can be regarded as concentrated at its centroid on the line AM. Hence we can think of the triangular plate as replaced by the bar AM whose density is propor- tional to the distance from the end A In accordance with Illustration 1, its centroid is at a point / two- FiG. 93. thirds of the way from A to M. The centroid of a triangle can also be located without any calculation whatever. From Fig. 93 it follows that the centroid lies on the median AM. The same argument shows that it lies on the medians BMi and CM2. Hence it lies at the point of intersection of the medians, i.e., at a point two-thirds of the way from a vertex to the middle of the opposite side. 4. The area of a semicircular plate of radius r. (Single integra- tion will be sufficient.) jl40] SUCCESSIVE INTEGRATION 271 5.1 Let OMKB, Fig. 94, be a quadrant of a circle of radius r. Let OMDB be a square. Denote by Ci, C2, and C3 the centers of gravity of the square, the quadrant of the circle, and the area MDBKM, respectively; and by Ai, A2, and A3 the corresponding areas. Then ^2X2 + AzX3 = AiXi A2X2 AiXi X3 = = 0.223 r. DC3 = 0.315 r. 6. A circular arc of radius r and central angle 2a. See Fig. 95. Ai \ / / \ \"~ /C2 *- \ V / -^^ D t i Fig. 94. Fig. 95. Hint. The centroid lies on the radius which bisects the central angle since this line is an axis of symmetry. Choose this radius as the axis of X and the center of the circle as the origin. Then y = 0, and - Xr! X ^ ~ Ira. ~ r cos Or do r'- — a ' Ira. /:.• cos do 2rc This problem can also be solved by using rectangular coordinates. Thus — t/r C( xdx Wj.2 — 2rc 2r^ sin a r sin a 2ra 7. The portion of the arc of the circle x^ + j/^ = r^ which lies in the 1 Exercises 5, 9, 20, 21, and 22 taken from Technical Mechanics by Maurer. 272 CALCULUS [§140 first quadrant. Use the result of Exercise 6. Also find the result directly. 8. The parabolic segment of altitude a and base h. See Fig. 96. Hint. Show that the equation of the parabola is ^ay"^ = b^x. Ans. X = fa. 9. A conical or pyramidal solid of altitude a and base A. Hint. Let OMNO, Fig. 97, represent the projection of the solid on the XF-plane. Divide the solid by planes parallel to the base into laminas or plates of thickness dx. Then the area of the lamina ' Y N ^.---^ ~" " < a >^> o f a ' K ^ / /^ \ ^^...^M \ ^--^^ \ , O ^ •"■""'''^ ! ! X dx Fig. 96. Fig. 97 , Ax"^ whose abscissa is x is — ^: and its volume is solid is -0-. Hence : Ax^dx The volume of the J'*'' /Ax^dx\ X = 1 Aa T 3a 4" Further the centroid of every lamina lies on the line joining the apex with the centroid of the base. Consequently the centroid of the solid lies on that line. 10. The hemisphere generated by revolving one quadrant of x» -(- j/2 = r^ about the A'-axis. Evidently 7/ = 2 = and ■I xy' dx 11. The surface of the hemisphere of Exercise 10. 12. The segment of a paraboloid of revolution of altitude h. §140] SUCCESSIVE INTEGRATION 273 13. The semi-ellipsoid of revolution generated by revolving one quadrant of -^ + r^ = 1 about the X-axis. 14. The surface of the paraboloid of Exercise 12. 15. The surface of a right circular cone. Any conical or pyramidal surface. 16. The area of the cycloid a; = a{d — sin e),y = a(l — cos 6). 17. The arc of the cycloid of Exercise 16. 18. The area in the first quadrant under x^ + y^ = a^. 19. The arc of the curve of Exercise 18 in the first quadrant. 20. The segments of the ellipse indicated in Fig. 98. It will be found that the centroid of the segment XAAX coincides with that of Fig. 98. the segment XaaX of the circumscribed circle, and that the centroid YBBY coincides with, that of the segment YhhY of the inscribed circle. 21. Ci and Cj are the centers of gravity of the two portions of Fig. 99. Show that their distances from the sides of the enclosing rec- tangle, a X b, are those marked in the figure. The curve OC is a parabola. See Exercise 5. 22. Find the centroid of the portion of a right circular cylinder shown in Fig. 100. C is the centroid. Its distance from the axis of the cylinder shown is — jr — -, and from the base is « H pj, — When the oblique top cuts the base in a diameter (lower part of Fig. 100) the distance of the centroid from the axis is -r^ and from the base 16 3ira ■32' 18 274 CALCULUS [§141 23. Find the centroid of the volume lying in the first octant and included between the cylinders x- + y^ = a^, x^ + z* = a^ 4«H \ C • 1 1 " "- 1 ■ > a __J_ ^^ ,C Fig. 99. Fig. 100. 141. Theorems of Pappus. Theorem I. The area oj the surface generated by revolving an arc of a plane curve about an axis in its plane and not intersecting it is equal to the length of the arc multi- plied by the length of the path described by its centroid. Theorem 11. The volume of the solid generated by revolving a. plane surface about an axis lying in its plane and not intersecting its boundary is equal to the area of the surface multiplied by the length of the path described by its centroid. Proof of I. Let ABC, Fig. 101, be an arc of length L lying in the XF-plane. Then y, the ordinate of its centroid, is given by the equation : H M Whence Fig. 101. / yds = y L. (1) (2) The surface generated by revolving the arc ABC about the X-axis is given by §141] SUCCESSIVE INTEGRATION 275 / S = 2Tr \ yds. (3) It follows then from (2) and (3) that S = 2TryL. (4) But 2iry is the length of the circular path described by the cen- troid of the arc ABC. Hence the theorem is proved. Proof of II. Let ABC, Fig. 102, be a plane surface of area A. Then y, the ordinate of its centroid, y is given by . / ydA -y = ^-j- (5) Whence ^ / ydA=Ay. (6) Fig. 102. Now the volume of the solid generated by the revolution of the area ABC about the X-axis is V = 2ir I j ydydx = 2t I y dA. (7) It follows from (6) and (7) that V = 2TrAy. (8) Hence the theorem is proved. • Exercises 1. Find the surface of the anchor ring generated by revolving the circle x'^ + {y — b)^ = a^, a < b, about the X-axis. 2. Find the volume of the anchor ring of Exercise 1. 3. Find, by using one of the theorems of Pappus, the centroid of a quadrant of a circular arc, radius a. Hint. The rotation of the arc about the X-axis, which coincides with a radius drawn to one extremity, generates the surface S = 2ira'^. Then, by (4), S = 27ra2 = 2T2/L = 2-Ky'^' 276 CALCULUS [§142 Hence y 2a 4. Find, by using one of the theorems of Pappus, the centroid of a quadrant of a circular area. 142. Centroid: Polar Coordinates. The formulas are readily obtained for finding the coordinates of the centroid of an area bounded by a curve whose equation is given in polar coordinates. The area of the element MPQR, Fig. 88, is p dp dd, and its moment about the F-axis is pdpdd p cos 6 = p"^ cos d dp dd. Hence Similarly, fCp"^ cose dp do ff pdpdd f fp^ sine dp dd ff pdpdd X = y = If it is advantageous, the integration with respect to d can be performed first. Exercises Find the coordinates of the center of gravity of: 1. The area of p = a(l + cos 6). The area of the upper half of the same cardioid. 2. The area of one loop of p = a cos 2d. 3. A circular sector of central angle 2 a. 4. One quadrant of a circle. A semicircle. (Obtain directly and also use the result of Exercise 3.) 5. The area of a portion of a cir- cular ring, Fig. 103, of radii R and r, and of central angle 2a. Denote by Cr the centroid of the sector of radius R, and by Cr that of the sector of radius r, and by C that of the given portion of the ring. Let the abscissas of these points be xr, Xt, and x, respectively, and let the corresponding areas be denoted by Ar, Ar, and A. Fig. 103. §143] SUCCESSIVE INTEGRATION 277 Then Hence Artr ■\- Ax = ArXr. - _ "^g^fl - ^rXr ^ 2 R^-r^ sin a ^~ A 3 R^-r^ a ' Obtain this directly by integration. 6. A segment of a circle of radius r cut off by a chord of length c. Use the method of Exercise 5. The distance of the centroid from the center is (.3 ^ 2r^ sin^ a 12A ~ M ' where A = area of the segment = r^ (2 a — sin 2 a:). 143. Moment of Inertia. Consider a system of masses, mi, rriz ■ ■ . , rrin, moving with linear accelerations, ji, j^, • • ■ , jn, respectively. The forces acting on these masses are then m\j\, m^ii, . . ., rrinjn, respectively; and the sum of the moments of these forces about an axis is equal to l^niijiri where ri, rz, . . . r„, respectively, are the moment arms of these forces with respect to this axis. If now the masses are rigidly connected and rotate about an axis, they have a common angular acceleration. Let the common angular acceleration be denoted by a. Then ^i = ari, ji = ar2, . . ., jn = oLTn, whcro ri, r2, . . ., r„ are the dis- tances of the masses mi, m^, . . ., rUn from the axis of rota- tion. The sum of the moments, T^mijiTi, becomes a'Zmiri^. This is the moment necessary to produce the angular acceleration a. To produce unit angular acceleration a moment equal to Sm.ri^ is necessary. This moment, ^miri"^, is called the moment of inertia, and is denoted by the symbol /. Thus I = SmiV. (1) The moment of inertia of the system would be unchanged if the n masses of the system were situated at a distance k from the axis of rotation such that 'Zniik'^ = 'ZmiTi^, or ^ ~ 2mi 278 CALCULUS [§143 k is called the radius of gyration. Its square is the mean of the squares of the distances ri, ri, . . . , r„ with respect to the mass. The moment of inertia of a system with respect to an axis of rotation plays the same role in the discussion of a motion of rota- tion as the total mass in the discussion of a motion of translation. In the former case the moment necessary to produce an angular acceleration a is a^niiri^. In the latter case the force necessary to produce a linear acceleration j is jSm,-. The kinetic energy of a rotating system can be expressed in terms of its moment of inertia and its angular velocity. If a particle of mass m is rotating with angular velocity co about an axis and at a distance r from it, its kinetic energy is equal to one-half the product of its mass by the square of its linear veloc- ity, i.e., to jWwV^. And if there is a system of particles of masses mi, m2, • • • , w„, at dis- tances ri, r2, . . ., r„, respec- tively, from the axis, all rotating with the angular velocity w, the kinetic energy of the system is equal to ^SmiCoVi^ = ^oo^Ziriiri^ = Wl- If a rectangular plate of uniform thickness ^ and composed of material of uniform density, p, rotate about an axis through one corner and perpendicular to its plane, its moment of inertia can be found by a process of double integration. Let the sides of the rec- tangle be a and h and take the origin at one corner. Fig. 104. The moment of inertia of the rectangle MNPQ is approximately the product of its mass, p^ Ay Ax, and the square of the approxi- mate distance, s/x^ -\- y"^, of its mass particles from the origin. That is, the moment of inertia of MNPQ is approximately p^(x2 -I- y^)Ay Ax. That of the strip EHIJ is approximately y = b ^^ P^ AySo DC-^' + y') AyAx=p^Ax i (x' + y') dy. And the moment of inertia of the entire plate is obtained by y J I Q P W M N X o B 7 J 1 Fig. 104. §143] SUCCESSIVE INTEGRATION 279 taking the limit of the sum of the moments of inertia of these strips as Ax approaches zero, viz., «/o Jo I = P^£to^^x\ {x^ + y')dy b (x2 + 2/2) dy dz = Wia-" + 6^), where M — p^ab, the mass of the plate. We have obtained the moment of inertia of the plate by inte- grating over its area the product of the mass of the element, p^ dy dx, by the square of its distance, ^Jx^ + y"^, from the axis of rotation. If instead of a rectangular plate we consider a plate of any shape, say that of Fig. 84, its moment of inertia is given by (x^ + y^) dy dx. (2) If the density, p, and the thickness, ^, are variable the foregoing argument shows that they must be written under the integral sign. For, the element of integration is p^{x^ + y"^) dy dx and only when p and ^ are constant can they be taken out from under the integral sign. If p and ^ are variable we have Pk {x^ + y^) dy dx. (3) (2) and (3) can be written in a form easily remembered, viz., I=Jr2dm, (4) where dm is an element of mass, and r is its distance from the axis. Sometimes in finding the moment of inertia of a body it is advantageous to choose the element of mass so that a single inte- gral will suffice. See for example Illustration 2, below. Illustration 1. Find / of a right-angled triangular plate whose thickness is 0.5 inch, and whose legs are 10 inches and 4 inches, about an axis through the vertex of the right angle and perpendicu- 280 CALCULUS [§144 lar to the plane of the triangle. The density of the material is 0.03 pound per cubic inch. See Fig. 105. I = 0.03-0.5 II (a;2 + y"^) dij dx. t/0 The student will carry out the integration and find the radius of gyration. Illustration 2. Find 7 of a circular plate about an axis through its center and perpendicular to its plane. . The plate has a radius of 10 inches. It is 2 inches thick and its density is 0.04 pound per cubic inch. Hint. Here it is convenient to divide the plate into concen- tric rings of inner radius r and of width dr. See Fig. 106. The Fig. 105. Fig. 106. volume of such a ring is 2-2Trr-dr, and its mass is 0.04*47rr-dr. The distance of this mass from the axis is r. Hence I = 0.16 ■s: dr. Also find the radius of gyration. 144. Transfer of Axes. Theorem. The moment of inertia of a body about any axis is equal to its moment of inertia about a parallel axis through the centroid, increased by the product of the jnass by the square of the distance between the axes. Let AB, Fig. 107, be the axis about which the moment of inertia is desired. Choose a system of rectangular axes such that the origin, 0, is at the centroid, such that the Z-axis is parallel to AB, §145] SUCCESSIVE INTEGRATION 281 and such that the ZF-plane contains the line AB. Consider an element of mass, dm, at P. The moment of inertia of the body about AB is then I = fiPBy dm = fUy - dy + x^l dm, or I = j {x"^ -\- y"^) dm — 2d j ij dm -{■ d^ j drn. (1) The first term of the right-hand side of (1) is the moment of inertia, /„, of the body about the Z-axis, an axis through the cen- troid. The second term, I y dm, is the moment of the body with respect to the XZ-plane, a plane passing through its centroid. Fig. 107. __ \y dm I dm Since y = 0, j y dm = 0. The last term, d"^ j dm, is d^M, where M is the mass of the body. Hence I = !„ + Md2. 145. Moment of Inertia of an Area. We have spoken of the center of gravity of an area quite apart from any idea of mass and have stated that this is a useful conception in the study of mechan- ics. In the same way the solution of some problems in mechanics requires the moment of inertia of an area quite apart from any idea of mass. 282 CALCULUS [§145 The moment of inertia of a plane area about an axis through the origin and perpendicular to its plane is defined by the integral. I = J J (x2 + y2) dy dx = Jf (x2 + y2) dx dy. The theorem on the transfer of axes holds in this case if the word "area" is substituted for the word "mass." Exercises Find the moment of inertia of the following : 1. A rectangle of sides a and b about one comer. (See Fig. 104.) About the centroid. About one base. About a line parallel to one base and passing through the centroid. 2. A right triangle, legs a and b, about one of the legs. About a line through the ^^ centroid parallel to this leg. 3. The area of a circle about an axis ' through a point on its circumference and perpendicular to its plane. See Illustration 2, §143. 4. The area of a circle about a diameter. Hint. I = X? 2xdy. Fig. 108. 5. The area of a circle about a tangent line. 6. The area between y = x^ and y^ = x about an axis through the origin perpen- dicular to the XF-plane. 7. A uniform bar of length L and linear density p about an axis through one end perpendicular to the bar. Find / about a parallel axis through the middle point of the bar. 8. A bar of length L, whose density is proportional to the distance from one end, about an axis perpendicular to the bar through the end of least density. 9. A slender uniform rod. Fig. 108, about a line through its middle point and making an angle a with the rod. Ans. I — ^s'lnL^ sin'' a, where m is the mass and L is the length of the rod. §146] SUCCESSIVE INTEGRATION 283 Hint. Denote by p the linear density. Then I = P i x^ sin^a dx, with proper limits. 10. The rod of Exercise 9 about a parallel axis through one end. 11. A wire bent into the form of a circular arc, Fig. 109, about the origin. Also find the moments of inertia, Ix and ly about the X- and F-axes, respectively. /- /- 7 = I r^rdB; Ix = I r^sin^erdd', /„ = I r^ cos^e r de. Fig. 109. 12. A triangle of base h and altitude h about an axis through the vertex parallel to the base. Divide the area into strips parallel to the base and of width dx. The axis of x is drawn from the vertex perpen- dicular to the base. _ , x^hx dx f 13. A triangle about a line through the center of gravity parallel to the base. Use the result of Exercise 12. 14. The area of the ellipse ^ + rj =1 about the major axis. (Use single integration.) About the minor axis. About the origin. 146. Moment of Inertia: Polar Coordinates. The moment of inertia of the element r dr dd about an axis through the origin is r^r dr dd. Hence the moment of inertia of an area is / = J Jr3 dr dd, with proper limits. If the moment of inertia of a plate is required, r' dr dd is to be multiplied by p, the density per unit area. Exercises Find the moment of inertia of the following : 1. The area of the cardioid p = a (1 + cos 6) about an axis through the origin perpendicular to the plane of the cardioid. About the initial line. 284 CALCULUS [§147 2. The area of one loop of p = o cos 2d about the initial line. 3. A circular sector of central angle 2a about the radius of symmetry. 4. The arc of the sector of Exercise 3 about the radius of sym- metry. 5. The area of Exercise 3, about an axis through the center of the circle and perpendicular to the plane of the sector. About a parallel axis through the centroid. 147. Moment of Inertia of a Solid. — We wish to find the moment of inertia of the solid of Fig. 86, about the Z-axis. The moment of inertia of the element of mass, p dz dy dx, about the Z-axis is equal to pix"^ + 7/^) dz dy dx. The total moment of inertia of the solid about this axis is the integral of this element throughout the solid. Hence, /. = I I I p(x2 + 7/2) dz dy dx. (1) Similarly r, - f r r J a Jy^ Jz = n"2 ni = fix. J Jz = Jo, Jy, Jz = y) /«= I j I p{y^ + z^) dz dy dx, (2) •z - fix, y) p{z^ + x2) dz dy dx. (3) If the solid be regarded as a geometrical volume of density 1, the p's disappear, and the formulas (1), (2), and (3) can be written /. = Jffi^' + y') dz dy dx, (4) I. = Jffiy' + 2') dz dy dx, (5) ly = ///(2' + x^) dz dy dx. (6) Let /„, = III px^ dz dy dx, (7) Lx = JJJpy'^ dz dy dx, (8) hv = fffp^^ dz dy dx. (9) The quantities (7), (8), and (9) will be called the moments of inertia of the solid with respect to the FZ-plane, the XZ-plane, and the XF-plane, respectively. They are the integrals of the §147] SUCCESSIVE INTEGRATION 285 product of the element of mass by the square of its distance from the respective planes. They can very frequently be found by a single integration by taking as element a plane lamina between two planes parallel to the plane with respect to which the moment is computed. If this is the case the moment of inertia about the coordinate axes can easily be found by noting that from the equa- tions(l), (2), (3); and (7), (8), (9): Iz = Iy^ + Ixz, I. = Ixz + Ixy, ly = Ixy + Iyz. That is, the moment of inertia about the Z-axis is equal to the sum of the mom£nts of inertia with respect to the YZ- and XZ-planes, and so on. In general, the moment of inertia of a body about an axis is equxil to the sum of its moments of inertia with respect to two perpendicular planes which intersect in that axis. In the same way it follows from the formula for the moment of inertia of an area, I = I I (a;^ + y^) dy dx, that the sum of the moments of inertia of an area {or a plate) about two perpendicular axes is equal to its moment of inertia about an axis perpendicular to the plane of these axes through their point of intersection. Illustration 1. Find the moment of inertia of the ellipsoid X V z —i + T^-{ — s = 1 about each of its axes. a^ 0^ c'- First Method. y /. = 8j j ^ a^-^'\^ a^ %J^ + Z^) dz dy dx. Carry out the integration far enough to see that it is not simple and then note the relative simplicity of the Second Method. Compute Ixz, the moment of inertia with respect to the XZ-plane. Take as element of integration the ellip- tical plate cut out by the planes y = y and y = y -\- Ay. The equation of the intersection of the ellipsoid and the plane y = y is x^ 2^ _ .»(i-fi) .<!-«;) 286 CALCULUS [§147 Now the area of an ellipse is t times the product of its semi- major and semi-minor axes. Hence the area of this ellipse is The volume of the elliptical plate in question is irac [l - p-j dy and its moment of inertia with respect to the XZ-plane is xacy^ (l - p-j dy. The total moment of inertia of the ellipsoid with respect to this plane is then I,, = Tracfy^[l-pjdy = 27roc ( /63 5 A 3 5/ 16 I XV can be written down at once as -. _ 47ra6c' Then T TIT ^Trahc ,, „ . We can write down at once by interchanging letters: Illustration 2. Find the moment of inertia of a right circular cone about a line through its vertex perpendicular to its axis, if the radius of the base is b and the altitude is h. Choose the vertex as origin and the axis of the cone as axis of x. Consider the plate bx of radius r cut out by the planes x = x and x = x + dx. Its §147] SUCCESSIVE INTEGRATION 287 moment of inertia about a diameter parallel to the axis of rotation through the vertex is equal to — j-rz — . (See expression for / of a circular plate about a diameter, Exercise 4, §145). Then the moment of inertia of this plate about the axis of rotation through the vertex is equal to this moment of inertia about an axis through the centroid increased by its volume (mass if density = 1) multi- plied by the square of the distance between this axis and the par- allel axis through the vertex (see §144) i.e., to Tb*x*dx irb^x'^dx ^ W "^ P ^ ■ And the total moment of inertia of the cone about the axis through the vertex is equal to the integral of this moment of inertia from X = to X = h. That is SX ■jrb^h , Trb%^ Trb-h ,, „ , ^,,. "20- +-5- = -20" (^^ + ^'^^)- Exercises Find the moment of inertia of: 1. The cone of Illustration 2, about a parallel axis through the centroid of the cone. About a diameter of the base. 2. A right circular cylinder, the radius of whose base is r, and whose altitude is h, about a diameter of one base. About a parallel axis through the centroid. 3. A rectangular parallelopiped with edges a, b, and c, about an axis through the centroid parallel to one edge. 4. A right circular cylinder about its axis. 5. A hollow right circular cylinder of outer radius R, inner radius r, and altitude h, about its central axis. About a diameter of one base. About a diameter of the plane section through the centroid perpen- dicular to the axis. 6. A right rectangular pyramid of base a X b and of altitude h, about an axis through the centroid parallel to the edge a. About an axis through the vertex and the center of gravity. Ans. /. = ^ (b- + Ih^). h=^ {a' + &=). 288 CALCULUS [§147 7. A frustum of a right cone about its axis if the radius of the large base is R, that of the small base is r, and the altitude is h. R^ — r^ Ans. I = -fifTrh -fz R — r 8. A hollow sphere about a diameter, if the outer radius is R and the inner radius is r. 9. A paraboloid of revolution, the radius of whose base is r and whose height is h, about the axis of revolution. Ans. I = lirr*h. 10. The anchor ring generated by revolving the circle [x - R]^ + y^ = r^ about the F-axis. Ans. h = TT^RrKR^ + fr"). ly = 27rmr\R^ + fr^). 11. A right circular cone about its axis. 12. A right elliptical cylinder of height L, and having the semi-major and semi-minor axes of its cross section equal to a and b, respectively, about an axis through the centroid parallel to b. 13. A quadrant of a circular plate about one of its bounding radii. 14. An equilateral triangle of side 2a, about a median. About a line through a vertex perpendicular to one of the sides through that vertex. CHAPTER XVI CURVATURE. EVOLUTES. ENVELOPES y 1 qI A- Pf /r+t^T o y / Fig. 110. 148. Curvature. Let PT and QT, Fig. 110, be tangents drawn to the curve APQ at the points P and Q, respectively. Denote the length of the arc PQ by As and the angles of inclination of PT and QT' to the positive direc- tion of the X-axis by t and t -j- Ar, respectively. At gives a rough measure of the deviation from a straight line, of that por- tion of the arc of the curve be- tween the points P and Q. The sharper the bending of the curve between the points P and Q the greater is At for equal values of As. The average curvature of the curve between the points P and Q is defined by the equation At Average Curvature = 't~'' (1) The average curvature of a curve between two points P and Q is the average change between these points, per unit length of arc, of the inclination to the X-axis of the tangent line to the curve. Or, more briefly, the average curvature is the average change per unit length of arc, in the inclination of the tangent line. The curvature at P is defined as the limit of the average curvature between the points Q and P as Q approaches P. On denoting the curvature by K, we have in accordance with the definition, ^' ^' (2) K = 1™ ■^ As = As ds The curvature at a point P is then the rate of change at this point of the inclination of the tangent line per unit length of arc. The 19 289 290 CALCITLUS I §151 curvature is a measure of the amount of bending of a curve in the vicinity of a point, 149. Curvature of a Circle. It is clear that the average curva- ture of a circle, Fig. Ill, is At _ At As r At Hence the average curvature is independent of As and conse- quently the curvature, the limit of the average curvature as As approaches zero, is A' = (1) Tfie curvature of a circle is constant and equal to the reciprocal of its radius. 150. Circle of Curvature. Radius of Curvature. Center of Curvature. Through any point P of a curve infinitely many cir- cles can be drawn which have a common tangent wdth the curve at P and whose centers are on the con- cave side of the curve. Of these circles there is one whose curvature is equal to that of the curve at P, i.e., one whose radius is equal to the reciprocal of the curvature at P. This circle is called the circle of curvature at the point P. The radius of this circle is called the radius of curvature, and its center the center of curvature, of the curve at the point P. The radius of curvature is denoted by R and, in accordance with (2), §148, its length is Fig. 111. R = K ds dr* (1) 151. Formulas for Curvature and Radius of Curvature: Rec- tangular Coordinates. For obtaining the curvature at any point on the curve y = /(x), we shall now develop a formula involving §151] CURVATURE. EVOLUTES. ENVELOPES 291 the first and second derivatives of y with respect to x. The above formula for curvature K can be written (1) dr dr dx dx dx \1 + [dx) Since . ,dy T = tan-^ -j-y dx d'y dr dx"^ Consequently and, by (1), §150, [^M^Q? , [-(S)r ^ - d2y dx2 (2) (3) We shall understand by K and R the numerical values of the right-hand members of (2) and (3), respectively, since we shall not be concerned with the algebraic signs of K and R. Illustration. Find the curvature oi y = x^. ^ = 2x ^ = 2 dx ' dx^ Substitution in formula (2) gives 2 K = (1 + 4x2)^ From this expression it is seen that the maximum curvature occurs when x is zero, and that the curvature decreases as x 292 CALCULUS [§152 increases in numerical value. When a; = 0, K = 2. When > — X -1-, -IV — 25 Exercises Find the curvature and radius of curvature of each of the curves: l.y = 2x- xK A. y = x^ — x^. 7. 2/ - 3x*. 2. 2/ = xi 5. y = -.• 8. y = x~^. -y-l- 6. 2/ = Vx. Vx 10. If p = /(&) is the equation of a curve in polar coordinates, show that „ ^ +Hd"gJ -^d¥^ K — , • Hint. ['-(^^)T _ dr _ d^ "~ ds ~ ds d« T = e +ip. (See Fig. 72.) til- _ d^ 50 ~ "^ do" Obtain -jr from the relation dff d0 IS given in de 152. Curvature: Parametric Equations. If the equation of a curve is expressed in parametric form, z = f{t), y = F{i), the cur- vature can be found by differentiating x and y and substituting in (2), §151. t can be eliminated from the result if desired. Illustration 1 . li x = t and y = t^, ^ ^ 2« and ^ = ^^ — = 2 dx ' dx' dt dx §153] CURVATURE, EVOLUTES. ENVELOPES 293 Hence 2 2 Illustration 2. Find the curvature of the ellipse x = a cos t, y = 6 sin t. dy b -J- = cot t. ax a d-hj b d .dt & „, 3-: = -V, cot t -3— = — csc^ t dx^ a at ax a r— T = iCSC^^ a sin t_\ a^ - -cscH „ a^ — ab — ab h. = [l+^'cot^f]^ (a'sinH + b^cosH)^ g ^2 + ^J ^^1 a*b* ib*x^ + a*y^Y Exercises 1. Find the curvature of the curve x = a cosh I, y — a sinh t. 2. Find the curvature of the curve x = a{t — sin 0) y = a{\ — cos 0- 153. Approximate Formula for the Curvature. If a curve dy deviates but little from a horizontal straight line, -r- is small and (dv\ 2 ^1 in formula (2), §151, is very small compared with 1. Hence the denominator differs very little from 1 and the formula becomes approximately ^=S- (') This approximate formula for K is frequently used in mechanics in the study of the flexure of beams. The slope of the elastic d'^y curve of a beam is so small that -v-^ can be used for the curva- ture without appreciable error. The approximate formula for the radius of curvature R is 7?- ^ ^~ d^ (2) dx^ 294 CALCULUS [§154 154. Center of Curvature. Evolute. Formulas will now be obtained for the coordinates of the center of curvature of a curve corresponding to any point P. Let the coordinates of P be a; and y. Denote by a and j8 the coordinates of the center of curvature of the curve at this point. There are four cases to be considered. See Fig. 112, a, b, c, d. Fig. 112. In Fig. 112, a, a = OM =^ ON - HP = X - Rsinr, /3 = MC = NP + HC = y + R cost. Since tanr = dy dx cost = NRS-f ; sinT dy dx /dy\^ \dx) §154] CURVATURE. EVOLUTES. ENVELOPES 295 Consequently, ^-m «=--f-d^f^' (1) and 1 + (^) ^ = y + — d^- (2) The student can show that, since -7— is negative for a descending curve and positive for an ascending curve, and since -r^ is positive when a curve is concave upward and negative when a curve is concave downward, formulas (1) and (2) hold for the three curves represented in Fig. 112, b, c, d. Illustration. Find the coordinates of the center of curvature corresponding to any point on the curve y = ± 2 y/~x. Only the positive sign will be used. If the negative sign is used it will only be necessary to change the sign of ^. dy ^ 1 dx y/^ ^ = _ J_ dx^ 2a;i* 1 1 1 + 2x1 = y 1 — = y - 2 V^(x + 1) = 2/ - 2/ [I + 1] = - 2x3 The equation of the locus of the center of curvature is obtained by eliminating x and y from the equations for a and /3 and the equation of the original curve. Thus 0= ~ 2 /.ox' X = — y— ; y = - (4)8)' . 296 CALCULUS [§155 Substituting in y"^ = 4a;, we obtain ^^ = A(« - 2)», the equation of the locus of the center of curvature. This is the equation of a semi-cubical parabola whose vertex is at the point (2,0). The locus of the center of curvature corresponding to points on a curve is called the evolute of that curve. Its equation is easily obtained in many cases by eliminating x and y from equations (1) and (2) and the equation of the original curve. Otherwise (1) and (2) constitute its parametric equations, a and /3 being ex- pressed in terms of the parameters x and y. Exercises 1. Find the evolute of t/ = 4x^. 2. Find the evolute of the ellipse I* ^ ^yi - 1 a^ "^ 6* ^' Hint It will be found that (a* - &2)x' (a« - 6*)y« « = ^, ; ^ = ^, Whence Elimination gives iaa)i + (6^)5 = (o2 - 62)5. 3. Find the parametric equations of the evolute of the cycloid, X = a{9 — sin 9), y = a(l — cos e). Ans. a = a(0 + sin 6), = — a(l — cos d). Show that the evo- lute is an equal cycloid with its cusp at the point ( — wa, — 2a). 4. Find the equation of the evolute of X = a(cos 9 + 9sm e), y = o(sin — ^cos^). Ans. a = a cos 9, /3 = a sin 9. Discuss. 155. Envelopes. If the equation of a curve contains a constant c, infinitely many curves can be obtained by assigning different values to c. Thus (x - c)2 + 2/2 = a2 (1) §155] CURVATURE. EVOLUTES. ENVELOPES 297 is the equation of a circle of radius a whose center is at (c, 0). By- assigning different values to c we get a system of equal circles whose centers lie on the X-axis. A constant such as c, to which infinitely many values are assigned, is called a parameter. A constant such as a, which is thought of as taking on only one value during the whole discussion, is called an absolute constant. We say that equation (1) represents Si family of circles or a system of circles corresponding to the parameter c. The general equation of a family of curves depending upon a single parameter can be written in the form, fix, y, c) = 0. (2) Exercises State the family of curves represented by the following equations contammg a parameter: 1. y = x^ + c, 2. y = mx + b, 3. y = mx + b, *• a» ^ &» ^' 6. 2/* = m(x + m), 6. x^ + y^ = a\ c being b being m being a being TO being a being the parameter, the parameter, the parameter. the parameter. the parameter, the parameter. Consider again the family of circles (1). Two circles of the family corresponding to the values, c and c -f- Ac, of the parameter intersect in the points Q and Q', Fig. 113. We seek the limiting positions of these points of in- tersection as Ac approaches zero. Clearly they are the points P and P', respectively, on the lines y = +a. Such a limiting posi- tion of the point of intersection of two circles of the family is called the point of intersection of two "consecutive" circles of the family. In general, the limiting position! f the point of intersec- tion of two curves, f{x, y, c), f (x, y,c-\- Ac), of a family, as Ac 298 CALCULUS [§155 approaches zero, is called the point of intersection of "consecu- tive" curves of the family. In the case of the family of circles (1) the locus of the points of intersection of "consecutive" circles is the pair of straight lines y = +a. This locus is called the envelope of the family of circles. In general, the envelope of a family of curves depending upon one parameter is the locus of the points of intersection of "consecutive" curves of the family. It will be shown in a later chapter that the envelope of a family of curves is tangent to every curve of the family. Exercise Draw a number of lines of the family X cos 6 + y sin d = p, where 9 is the parameter, and sketch the envelope. A general method of obtaining the envelope of a family of curves will now be given. The equation of a curve of the family is f(x, y, c) = 0, (3) where c has any fixed value. The envelope is the locus of the limiting position of the point of intersection of any curve (3) of the family with a neighboring curve, such as fix, y, c + Ac) = 0, (4) as the second curve is made to approach the first by letting Ac approach zero. The coordinates of the points of intersection of the curves representing equations (3) and (4) satisfy fix, y, c + Ac) - fix, y, c) = 0. (5) Then they satisfy fix,y,c + Ac) -fix,y,c) Ac = 0, (6) since Ac does not depend on either x or y. Then the coordinates of the limiting positions of these points of intersection satisfy lim fix,y,c + Ac) -fix,y,c) _ Ac A ^c ~ §155] CURVATURE. EVOLUTES. ENVELOPES 299 The first member of this equation is the derivative of /(x, y, c) with respect to c. It may be written in the form, df(x, y, c) _ , dc ~ ^- ^'' The differentiation is performed with respect to c, x and y being treated as constants. The point of intersection also lies on (3). Hence the equation of its locus is obtained by eliminating c between (3) and (7). Illustration 1. Find the equation of the envelope of the family of circles, {x — cy -{- y^ = a^, c being the parameter. The equation of the curve written in the form f{x, y, c) = is {x - cY + 2/2 _ a2 = 0. (I) Differentiating with respect to c, - 2{x - c) = 0. (II) The elimination of c between (I) and (II) gives or y = ±a, as the envelope. Illustration 2. Find the equation of the envelope of the family of lines, x cos 6 -^ y sin d = p, 6 being the parameter. On differentiating the first member of X cos d -\- y sin d '- p = (I) with respect to d we obtain — X sin 6 -\- y cos d = Q. (II) The result of eliminating d between (I) and (II) is ^2 _j_ ^2 _ p2^ a circle of radius p about the origin as center. Exercises P 1. Find the envelope of the family of straight lines y — mx -\ — ' m where m is the parameter. Draw figure. 300 CALCULUS [§156 2. Find the envelope of the family of lines y = mx + o\/l + m', where m is the parameter. Draw figure. 3. Find the envelope of the family of parabolas y^ = c(x — c), c being the parameter. 4. Find the envelope of the family of lines of constant length whose extremities lie in two perpendicular lines. 6. Find the envelope of y = px — -p^, p being the parameter. Draw figure. 6. Find the envelope of the family of curves (x — c)^ + 2/^ = 4pc, c being the parameter. Draw figure. 7. The equation of the path of a projectile fired with an initial velocity Vo which makes an angle a with the horizontal, is y — X tan a — n — ; 7. — " 2v^ cos^a Find the envelope of the family of paths obtained by considering a a parameter. Wo* gx^ Ans. V == -ji— — 77—! 8. The equation of the normal to y^ = 4x at the point P, whose coordinates are Xi and yi, is yi , ■. y - yi = - 2^ (x— xi). Since yi^ = 4xi, this may be written 2/iX +2y -'^ 2yi= 0. Find the equation of the envelope oi the normals as P moves along the curve. Hint. On differentiating with respect to the parameter 2/1 we obtain a. o V^^^Z? On substituting this value of yi in the equation of the normal and squaring we obtain 2 4(x - 2)^ y = —27 — This is the evolute of the parabola as we have seen in §164. 156. The Evolute as the Envelope of the Normals. In Exercise 8 of §155 it was seen that the evolute of a parabola is the envelope of its normals. This is true for any curve. The result is fairly evident from the examination of the curves of the exercises of §156] CURVATURE. EVOLUTES. ENVELOPES 301 §165 and their evolutes. It will be shown that the normals to a curve are tangent to its evolute. The parametric equations of the evolute are a = X — Rsinr, (1) ^ = y + RcosT. (2) On differentiating with respect to the variable s, which is per- missible since x, y, R, and r are aU functions of s, we obtain da dx dR . _, dr 3— = 3 5— sm T — /t cos r 3- • ds ds ds ds d^ dy , dR „ . dr T" = 3 r T' COST — R smr -j-- ds ds ds ds Now = COST, dx ds dy ds ~ R' Then the foregoing equations become da dR . ds ds ' d^ dR 3— = 3— COS T. ds ds Hence the slope of the tangent to the evolute is ^ = -cotT (3) da Therefore the tangent to the evolute is parallel to the normal to the curve at the point (x, y) to which (a, jS) corresponds. But the normal to the curve at (x, y) passes through (a, /3). Hence it is tangent to the evolute at this point. It can also be shown that if Ci and C2, Fig. 114, are the centers of curvature corresponding to the points Pi and P2, the length of the arc C1C2 of the evolute is equal to the difference in the lengths of the radii of curvature, Ri and R^. For, from the above values of da and d^ it follows that Vda^ + diS* = dR. 302 CALCULUS [§157 But VdoM-d^ is the differential of the arc of the evolute. Call it d(T. Then da = dR, and hence on integrating a = R -\- C. a and R are functions of s, the arc of the given curve. Then corre- sponding to a change As( = arc P1P2) in s, cr and i2 will take on the increments Ac and AR which are equal by the foregoing equation. But A(T = arc C1C2, and AR = R2 — Ri. Hence arc C1C2 equals ^2 — Ri- Fig. 114. 157. Involutes. In Fig. 114, suppose that one end of a string is fastened at C and that it is stretched along the curve CdCiKM. If now the string be unwound, always being kept taut, the point M will, in accordance with the properties of the evolute, trace out the curve MP1P2P. This curve is called the involute of the curve KC1C2C. If longer or shorter lengths of string, such as CKM2 be used, other involutes will be traced. In fact to a given curve there correspond infinitely many involutes. The given curve is the evolute of each of these involutes. We see that while a given curve has but one evolute it has infinitely many involutes. In Exercise 4, §154, the circle x = a cos 6, y = asin 6 was found as the evolute of the curve x = a(cos 6 -\- 6 sin 6), y = a(sin 6 — 6 cos 6). Then the latter curve is an involute of the circle. The student will draw a figure showing a position of the string as it would be unwound to generate the involute and indicate the angle 6. CHAPTER XVII SERIES. TAYLOR'S AND MACLAURIN'S THEOREMS. INDETERMINATE FORMS. 158. Infinite Series. The expression Ui + U2-\- U3-\- ■ ■ ■ + Un+ ■ ' ' (1) where ui, Uz, "Wa, • • •, %i„, ' • ' is an unlimited succession of numbers, is called an infinite series. Let s„ denote the sum of the first n terms of the infinite series (1). Thus Sn = Ul + U2 + U3 + • • • -\- u„. (2) If, as n increases without limit, Sn approaches a limit s, this limit is called the sum of the infinite series, and the series is said to be convergent. Illustration 1, In Illustration (1), §21, AB, Fig. 21, is a line 2 units long. The lengths Axi, X1X2, XiXz, x^Xt, • • • x„_ix„, • • • are 'i-, h, l, h ' ' ' > n~i t ' ' ' > respectively. For this series Sn = 1 + ^ + ^ + I + • • • + -^f (3) The limit of this sum as n increases without limit is 2, as the figure shows. Or, we may write, l + h + l + l+ • • • +2^,+ • ■ ' =2. (4) Illustration 2. The sum of the geometrical progression 1 + r + r^ + r^ + r" + • • • + r" (5) does not approach a limit if \r\ y 1, but if \r\ < 1 it approaches the limit z , when n becomes infinite. 1 — r' The infinite series (1) is said to be divergent or to diverge if, as n increases without limit, s„ does not approach a limit. 303 304 CALCULUS [§158 Thus the series l+2 + 3 + 4+---+n + ---, l-1 + l-l + l-l + l---, • 1 + R-\- R^-\- R^ + R*+ • ■ • + R^+ • ■ ' {R> I) are illustrations of divergent series. If the terms of an infinite series are functions of a variable x and if the series is convergent for any range of values for x, the series defines a function of x for that range of values. Thus, if |x| < 1, the series 1 +x + x^ + x^ + • • • +x^ + • ' ' (7) defines the function z . On the other hand, if the series is 1 —x ' divergent it does not define a function of x. Thus, if \x\ > 1, the series (7) is divergent and does not define the function ^ _ ' or any other function. It may happen that the sum of a few terms of an infinite series representing a function is a very close approximation to the value of the function. As an illustration take the infinite geometrical progression (7), which when |x| < 1 represents the function . _ . If the terms after x*~^ are neglected, the error is X* + x*+i + • • . + X" + • • • = r^—- 1 — X X* The error, :j-^ — is very small compared with the value of the function, q , and decreases as k increases; i.e., a better and 1 — X ' ' better approximation is obtained the greater the number of terms retained. Another infinite series is obtained by expanding (1 + x)' by the binomial theorem, (1 + x)' = 1 + -|x + Ix' - -Ax^ + • • • (8) This series can be shown to be convergent when |x| < 1 and divergent when |x| > 1. Just as the function . _ is represented to a high degree of ap- proximation by the first few terms of the series (7) when jxj is small, §159] TAYLOR'S AND MACLAURIN'S THEOREMS 305 the function {I ■}- x)^ is represented approximately by the first few terms of (8) when |a;| is small. In both cases the functions are represented approximately by polynomials. A method will be developed in the succeeding articles which will enable us to deter- mine polynomial approximations to other functions, such as sin x, tan X, e'. An infinite series of the form ao+ di'X + o.-ix''' + asx' + . . . + OnX" + . . . (9) is called a 'power series in x. One of the form Oo + ai(a; — a) + a^ix — ay + 03(3; — a)' + • • • + an{x - a)" + • • • (10) is called a power series in (x — a). The series (7) and (8) are power series in x representing the functions ^ _ and (1 + x)^, respectively. In the succeeding articles power series will be obtained representing the functions sin x, tan x, e*, etc. 159. Rolle's Theorem. Let f{x) be a single-valued continuous function between x = a and x = b, having a continuous first Fig. 116. Fig. 117. Fig. 118. derivative, f'{x), between the same limits. Further, let /(o) = and f(b) = 0, i.e., let the curve representing the function cross or touch the X-axis at a; = a and x = b. The curve may or may not ?o 306 CALCULUS [§160 cross or touch the X-axis at intermediate points. (See Fig. 115.) Since f{x) is continuous it cannot have a vertical asymptote be- tween X = a and x = b as shown in Fig. 116, nor can it have a finite discontinuity as shown in Fig. 117. Since f'{x) is con- tinuous between x = a and x = h, the curve cannot change its direction abruptly between these limits, as shown in Figs. 118 and 119. Since cases such as are represented by Figs. 116, 117, 118, and 119 are excluded, the curve. Fig. 115, must have a horizontal tangent at some point x = Xi between x = a and x = b. Hence the Fig. 119. Fig. 120. Theorem. If f{x) is a single-valued function from a; = a to X = b, andiff(x) andf'(x) are continuous between these limits, and further if f{a) = andf(b) = 0, thenf'ixi) = 0, where a < Xi < b. 160. Law of the Mean. Let f(x) be a single-valued function between x = a and x = b. Further let f{x) and f'{x) be con- tinuous between these limits. Fig. 120. It is then apparent from the figure that at some point P between A and B, the tangent line to the curve will be parallel to the secant line AB. Hence the Theorem. If f{x) and f'{x) are continuous between x = a and X = b, then J ^""'^ - b-a ' where a < Xi < b, or fib) = /(a) + (& - a)f(x^). (1) An analytic proof of this law will also be given. Define a number Si by the equation fib) = fia) + ib- a)Si, or fib) -fia) -(6-a)>Si = 0. (2) §161] TAYLOR'S AND MACLAURIN'S THEOREMS 307 It will be shown that Si = f'{xi), where a < Xi < b. From the first member of (2) build up the function 0i(a;) by replacing a by X. <l>iix) =m -fix) - {b-x)Su (3) Then <f>'i{x) = -f{x)+Su (4) Since f{x) and f'(x) are continuous between x = a and x = b, 0i(a;) and <f>'i(x) are continuous between the same limits. By (3) and (2), <^i(a) = 0, and by (3), <t>i{b) = 0. Hence <^i(x) satisfies the conditions of RoUe's Theorem and consequently <l>'i(xi) = 0, or /'(xi) - 5i = 0, or Si^f'ixi), where a < xi < b. On substituting this value of Si in (2) we obtain m =/(a) + (6-a)/'(xi), which proves the theorem. 161. The Extended Law of the Mean. Let f(x) be a function which with its first and second derivatives, f'{x) and /"(x), is continuous from x = a to x ^ b. Define a number Sz by the equation m = f{a) + {b- a)f'{a) + ^-^-^S^, (1) or m - /(a) - (6 - a)f'ia) - ^^^^S, = 0. From the first member of the latter equation, form the func- tion (f>2{x) by replacing a by x: Mx) = m - /(x) - (6 - x)/'(x) - ^^^'-5,. (2) Then 0'2(x) = -fix) - (6 - .x)/"(x) +/'(x) + (& - x)S, = {b-x) \S^ - /"(x)]. Since /(x), /'(x), and/"(x) are continuous, ^lix) and 4>\{x) are continuous. Further by (2) and (1), 02(a) = 0, and by (2), 308 CALCULUS [§162 02(&) = 0. Hence the conditions of RoUe's Theorem are satisfied, and <t>'2{x,) = 0, (3) where a < X2 < h. Or [h - X.MS2 - fix,)] = 0, or ^2=r(x2). (4) On substituting this value of Sz in equation (1), we obtain m = /(a) + (6 - a)na) + ^^^^fix,), (5) where a < x^ <. h. 162. Taylor's Theorem with the Remainder. Finally let j{x) and its first n derivatives be continuous from x = a to x =b. Define Sn by the equation m = f{a) + (6 - a)r{a) + ^^^^f{a) + ■ • • m - m -ib- a)f'{a) - ^^-—^fia) - • • • I n — 1 "^ I n Form the function </)„(x) by replacing a in the last equation by x. *.(x) =./(6) -/(x) - (6 - i)/'(x) - ^^-=^V(^) - ■ • ■ _((^)^ (6-x)« n — 1"^ _ri Then <^ "C^) = - |n_i /^"^(^) + |^_i 'S:,. (3) Since /(x),/'(x), • • •, /W(x) are continuous, <^„(x) and </)'„(a;) are continuous. By (2) and (1) <i)n{a) = 0, and by (2) 4>n{h) = 0. Hence the conditions of Rolle's Theorem are satisfied, and 'P'niXn) = 0, §162] TAYLOR'S AND MACLAURIN'S THEOREMS 309 or S„=f(-^{Xn) (4) where a < x„ < 6. Hence the Theorem: If f{x), f{x), f"{x), • • •, /("Ka;) are continuous from X = a to X = b, m = f(a) + (b- a)!' (a) + ^^^^ r(a) + ■ ■ ■ where a <. x„ <. b. This theorem, which is only an extension of the theorem express- ing the law of the mean, is called Tmjlor's Theorem with the remainder. The last term is called the remainder. If b is replaced by x, (5) becomes /(x) = /(a) + (X - a)}\a.) + ^"^'/'(o) + • ■ - where a < x„ < x. This inequality is sometimes written Xn = a -\- e{x - a), where < B < I. Illustration 1. Let/(x) = e*. Then /(x) = e' f{a) = e» fix) = e» /'(o) = e» /"(x) = e' /"(a) = e- /(")(a;) = e* f^"\a) = e- Hence by (6) l + (x-a)+-^"-^+ • • • 4-(^-«)"-^ In - 1 + (^^^,... (7) If a = 0, If a = and x = 1, «^2 /j«7l ~" 1 ']*'* e=l + l + ,^+. • •+r^ + T^«-- (9) 310 CALCULUS [§162 The remainder in (8) and (9) can be made as small as we please by choosing n sufficiently large. Taylor's Theorem may be expressed in still another form by setting b in (5) equal to a -\- h. fia + h)= Si.a) + hna) + ^ /"(a) + • • • + ^/^"^(xn) (10) where a < x„ < a + A, or Xn = a + 0/i, < ^ < 1. If the values of a function and its derivatives are known at a, then the values of the function at a point a -\- h can be computed by this formula. In (10), /(a + A) is represented approximately by a poly- nomial of degree n — \ in h. The coefficients are the derivatives of /(x) at X = a. The error in the approximation is given by the last term. This term gives only a means of estimating the error, since Xn is not known. The maximum error can, however, be determined by substituting M^^\ the greatest numerical value of /^">(a;) in the interval (a, a -\- h) for f''\xr.). The numerical value of the error is therefore less than |n If a = 0, (6) becomes f(x) = f(0) +f (0) X + f"(0) % + ' • • + f^"~'' (0) ^5^^ 12 l(n-l) + f^"nx»)^» (11) where < Xn <• x, ov Xn = 6x, < d < 1. In (11) it is assumed that the function f{x) and its first n derivatives are continuous from x = to x = a. (11) is known as Maclaurin's Theorem with the remainder. Illustration 2. Expand sin x by Maclaurin's Theorem in powers of z as far as the term containing x®. /(x) = sin X /(O) = f{x) = cos X /'(O) = 1 /"(x) = - sin X /"(O) = fix) = - cos X /'"(O) = - 1 /■^(x) = sin X rxo) = P{x) = cos X /"(O) = 1 /^*(x) = — sin X rm = /^"(x) = - cos X r'\x.) = — COS X7. §163] TAYLOR'S AND MACLAURIN'S THEOREMS 311 Substitution in (11) gives X'' sina; = x — ro+j-g— i^ cos Xi, (12) where <. Xt <. x. Since |cos X7I < 1, sin x differs from x^ x^ ^~ [3 + j5 x'' by a number less than r«* In general, since sin x and its derivatives are continuous, x^ , x^ x" , , ^ , s smx = X — r^ + re ~ ' ' "if" (sin x„ or cos x,,), (13) where < x„ <. x. Thus the difference between sin x and ^ ~ [3 + [5 ~ ■ ■ " - \r^^^ x" is less than r—, a number which for a given x can be made as small \n as we please by taking n sufficiently large. Hence the series (13) can be used in computing the value of sin x. If x is small, only a few terms of the series need be used to obtain a very close ap- proximation to sin X. Thus in formulas in which sin x occurs, sin X is frequently replaced by x if the angle is small. Such a sub- stitution was made in equation 1, §81. It must be remembered in making the substitution that x is expressed in radians. 163. Taylor's and Maclaurin's Series. If f{x) and all of its derivatives are continuous within an interval, the number of terms in (6), (10), and (11), §162, can be increased indefinitely. These equations then become, respectively, f (x) = f(a) + f (a) (X - a) + f"(a) ^^^' + • • • + f^"Ha)^^^%---. (1) f(a+h)=f(a) + f'(a)h + r(a)j'^+- • • + f (">(a)^+ • • •. (2) f(x)=f(0) +f (0)x+ f"(0)^^ + • • • + f ^"^ (0)^ + • • •• (3) 312 CALCULUS [§163 In (1), f{x) and its derivatives are assumed to be continuous from a to x. In (2) , /(x) and its derivatives are assumed to be continuous from aio a + h. In (3), /(x) and its derivatives are assumed to be continuous from to X. The series (1) and (2) are called Taylor's Series and (3) is called Maclaurin's Series. If we denote the last term in each of the equations (6), (10), and (11), §162, by Rn, it is necessary that i^h^n = in order that (1), (2), and (3) shall represent /(x), /(o + h), and /(x), respectively. Such series represent a function only so long as they are conver- gent. Later in this chapter means of testing the convergence of series will be discussed. The series (1), (2), and (3), if convergent, represent /(x) but do not give a means of estimating the error made by stopping with a given term. This can best be deter- mined from the expression for the remainder 72„ in Taylor's or Maclaurin's Theorem with the remainder. Illustration 1. Represent sin x by a power series in (x — a). Use formula (1). /(x) = sin X /(a) = sin a /'(x) = cos X f'ia) = cos a j" (x) = — sin X /"(o) = — sin a. /'"(x) = — cos X /'"(«) = — cos a fix) = sin X /'^(«) = sin a P{x) = cos X /^(o) = cos a Then by (1) , , . . (x- ay (x - ay sin X = sin a + cos a {x — n) — sm a — r^ cos a — r^ — . (x - a)* {x -ay + sm a — Hi + cos a , - — - — • • •. The corresponding Maclaurin's Series is obtained by letting a = 0. x' , x^ x^ , sin X = x - rg + , 5 - 1^ + • . •. §164] TAYLOR'S AND MACLAURIN'S THEOREMS 313 Illustration 2. Expand tan x in a power series in x. fix) = tan X. /(O) = fix) = sec^x. f'iO) = 1 fix) = 2 sec^x tan x. /"(O) = = 2(tana; + tan'x). fix) = 2(sec2a; + 3 tan^x sec^a;) /'"(O) = 2 = 2(1 + 4 tan^x + 3 tan%) fix) = 16 tan X sec^x + 24 tan^x sec^x /"'(O) = = 16 tan X + 40 tan^x + 24 tan^x fix) = 16 sec^x + 120 tan^x sec^x + 120 tan^x sec^x fiO) = 16 On substituting in (3) we obtain X' 2x« tan x = x + -^ + Yk''t~'' '• The next two terms are ^W ^^ and ^H^ x". Exercises 1. Expand cos x in a power series in x. 2. Expand cos x in a power series in (x — a). 3. Expand cos (a+ ^0 in a power series in h. 4. Expand sin (a + A) in a power series in h. 6. Express the remainder after three terms in each of the series of Exercises 1, 2, 3, 4. 6. Expand e" in a power series in x. 7. Expand e"'^'^ in a power series in h. 8. Expand e" in a power series in {x — a). 9. Expand log (1 + x) in a power series in x. 10. Expand log (1 — x) in a power series in x. 11. Expand tan'^x in a power series in x. 12. By the use of the series already found, compute: (a) ^e to 5 decimal places. (6) \^e to 6 decimal places, (c) sin 3° to 6 decimal places. (d) cosine of 1 radian to 4 decimal places. 13. By the use of the result of Exercise 3, find cos 33° correct to 4 decimal places. 14. By the use of the result of Exercise 4, find sin 32° correct to 4 decimal places. 164. Second Proof for Taylor's and Maclaurin's Series. These 314 . CALCULUS [§164 series can be obtained very simply in another way if we make certain assumptions and do not attempt to justify them. Assume that j{x) can be represented by an infinite power series in (x — a): f{x) = aa-\-ai{x—a)-\-a2{x—aY+' • • +a„(a;— a)"+ • • •, (1) where Oo, Oi, 02, ' ' ', On, ' ' ' are coefficients which are to be determined. Assume further that the result of differentiating the second member term by term any given number of times, is equal to the corresponding derivative of the first member. Then, f'{x) = ai + 2a2(x — a) -\- 803(0; — a)^ + • • • + nan(x — a)"~^ + • ■ fix) = 202 + Ga^ix - a) + ■ ' ■+ n{n - l)a„(a; - a)--^ + ■ ■ ■ (2) f"'ix) = 603 + • . • + n(n - l)(n - 2)a„(x - a)"-^ +• • /(") (x) = |wo„ + . . . Put re = a in (1) and (2). whence /(a) = ao, /'(a) = ai, r(a) = 2a2, /'"(a) = I3a3, /"Ha) = Inan. ao = /(a), ai = f'(a), fia) a2 = ^v at = —f^ — > /("^ («) Ctn = , > §165] TAYLOR'S AND MACLAURIN'S THEOREMS 315 Substituting in (1) we obtain fix) = f{a) + /'(a)(x -a)+ -^(x - a)' By setting a = 0, and x = a + h,we get (3) and (2), respectively, of the preceding section. 165. Tests for the Convergence of Series. Several tests will now be given for determining whether or not a series is convergent. They will be given without proof, though in most cases the proof is not difficult. If a series Ui -\- uz -{• • • • + Wn + • • -is convergent, lina Un = 0. n— 00 The converse of this statement is not true. Thus the series 1 + i + i + i + i + • • • + ^ + • • • (1) is divergent, although ^^"^ - = 0. n=oo^ That this series is divergent can easily be seen as follows: 3 T^ 4 -^ 2 "5 + C + Y + "8 > 2 « ~r 1 0' + 1 f -P "1 2" "T 1 3' + "1 4" "T I 5 "1" 1 6" ^ 2 The terms of the series can then be grouped into infinitely many groups such that the sum of the terms in each group is greater than 5. But the series is divergent. Much more then is the series (1) divergent. Test 1. If ^^"^ Un is not zero the series is divergent. This test n= <x> is easy to apply and if it shows the series to be divergent, no further investigation is necessary. 316 CALCULUS f§166 Test 2. Alternating series. A series of decreasing terms whose signs are alternately plus and minus and for which n— 00 is convergent. Thus the series l-l + l-\ + l-ls+ • • ' (2) is convergent. The reason for the convergence of such an alternating series can be seen as follows. Denote by Sn the sum of the first n terms and suppose the (n + 1)**^ term positive. (See Fig. 12L) Then, since the terms are constantly decreasing, Sn + l > Sn', Sn+2 <^ Sn + l', Sn+2 > Sn- 'Sfl+l ' '\^ Sn+2 -*{ Fig. 121. It is clear that as n increases Sn oscillates back and forth but always within narrower and narrower limits, owing to the fact that the terms are constantly decreasing. As n becomes infinite the amount of this oscillation approaches zero since 71= 00 Sn therefore approaches a limit. Test 3. Comparison Test. If the terms of a series are in numer- ical value less than or equal to the corresponding terms of a known convergent series of positive terms, the series is convergent. If the terms of a series of positive terms are greater than or equal to the corresponding terms of a divergent series of positive terms, the series is divergent. A useful series for comparison is the geometrical series a -\- ar -^ ar^ + ar'' + ■ • • + ar" + * * • , (3) which is convergent if |r| < 1 and divergent if |r| ^ L See also the series (a) of Illustration 2 of this section. §165] TAYLOR'S AND MACLAURIN'S THEOREMS 317 Test 4. The Ratio Test. By comparison with the geometrical series it can be shown that the series is convergent if divergent if If lim Un+l n=a> Un lim Un+l n=co Un lim Un+l «=« Un <1, > 1. = 1 the test fails. In this case other tests must be appHed. There are a great many tests for the convergence of series but only a few can be given here. It should be added that there is no test that can be applied to all cases. Illustration 1. Test the series 1 - t+l-i + + for convergence. Since lii Un is not zero, the series is divergent (Test 1). It is to be noted that the terms of the series are alternately positive and negative and that they decrease, but they decrease to the limiting value 1 instead of 0. Hence test 2 does not apply. Illustration 2. Test the series 1 + -+-+-" + ^ ~ 9< ' *?« ~ /!< ' (a) for convergence. This series is useful in testing the convergence of series by comparison. If f = 1 we have seen that this series is divergent. (See (1).) If f < 1 each term of (a) is greater than the corresponding term of (1) and hence (o) is divergent. If f > 1 we can compare (a) with l + l + l + l, + i + ll + l.+ "• (« 318 CALCULUS r§165 Each term of (a) is less than or equal to the corresponding term of (6). But (6) is convergent since it can be written ^ + ^(^)+M4^)+M^)-^ 1+-+-+^+-+ which is a geometric series whose ratio, ^~, is less than 1. Hence (a) is convergent when t > 1. Summing up: (a) is divergent if t ^ 1. (a) is convergent U t > 1. Illustration 3. Test the series 1 1 i + r2 + r3+i4 + - • • for convergence. Apply test 4. 1 Mn = P- jn Hence 1 lim Un+1 lim 1^ + 1 lim 1 M=oo Un 1 "=- n + 1 In = 0. The series is therefore convergent. Illiistration 4. For what values of x, if any, is the series ^~ 13"^ 15 17 + convergent? Then 2n -1 lim n= 00 Un+l Un a;2„+i im |2n + l = QO a;2„-i |2n-l lim n = oo |2n (2n + 1) = 0, §165] TAYLOR'S AND MACLAURIN'S THEOREMS 319 for all finite values of x. Hence the series is convergent for all finite values of x, positive or negative. Illustration 5. For what values of x is X'' '^3 X* ,x. convergent? \Un\ = 71 _ lim ^n + l = lim n= 00 X Un + l n+ 1 n Un n + 1 n lim n=oo The series is therefore convergent if |x| < 1. Furthermore it is convergent if x = 1 (Test 2), and divergent if x = — 1. (See series (1).) As has been stated there is no one test of convergence which can be applied with certainty of success to any given series. The tests which can be most frequently applied have been given. It is suggested that the following procedure be observed in general. 1. See if ^^^Un = 0. n= 00 2. If so, is test 2 applicable? 3. Tf not, try the fatio test, test 4. This will fail if lim n= 00 M„ 1. 4. In this case, and in cases where the other tests fail or are difficult, try the comparison test. Exercises Test the following scries for convergence. \. \ -\+\ -I + i\ -■ • •■ 2. i - I + -f - I + A - • • •• 3. 4 + i + ^ + ^ + A + • • •. |3 |4 |5 10* ^ ^ A A 13 + 14+ I 5 + ' ' *• 5. 10 "*" 10^ "^ 10' 1 12 + 320 CALCULUS [§166 ^- ^ + 2V2 + 3V3 +4V4 + " ' ■• ' 32 ^ 52 72 ^9=* ^' 1-2 ^ 3-4 ^ 5-6 ^ For what values of x are the following series convergent? 10. The Maclaurin's series for e"? Exercise 6, §163. 11. The Maclaurin's series for cos x? Exercise 1, §163. 12. The Maclaurin's series for sin x? Illustration 1, §163. 13. The Maclaurin's series for log (1 — x)? Exercise 10, §163. 14. The Maclaurin's series for tan~^x? Exercise 11, §163. 166. Computation of Logarithms. The series of Exercise 9, §163, for log (1 + x) is convergent only when —1 <x^ + 1, and that for log (1 — x), Exercise 10, §163, only when — 1 ^x< + 1. It would appear then impossible to find the logarithm of a number greater than 2 by these formulas. By a very simple device it is, however, possible to obtain formulas for finding the logarithm of any number. From the series of Exercises 9 and 10, §163 it follows that 1 -f J log:j = log (1 + x) - log (1 - x) X+3-+g-+---], (1) where Ixl < 1. Let x = ^ — i~T' Then ' ' 2z + 1 1 -hx _ z+ 1 1 — X z z where > 0, or log(^+l)=log3+2[2^+3(22 + l)3+5(22Vl)^+' '^^^ By letting z = 1, log 2 can be computed by this formula. The series is much more rapidly convergent than that for log (1 -\- x), X = 1. In fact, 100 terms of the latter series must be taken to and ^°^ ~T~ ^ ^L2M^ "^ 3{2z + iy "^ 5(27+Tr^ + • • -J . (2) §168] TAYLOR'S AND MACLAURIN'S THEOREMS 321 obtain log 2 correct to two decimal places, while four terms of the new series (3) will give log 2 correct to four decimal places. After log 2 has been found, log 3 can be found by setting z = 2. The logarithm of 4 is found by taking twice log 2; log 5 by setting 2 = 4; log 6 by adding log 3 and log 2, and so on. Exercise Compute log 5 correct to four decimal places, given that log 4 = 1.38629. Here, as always in the Calculus, the base is understood to be e. 167. Computation of tt. By letting a; = 1 in the series for tan~'x. Exercise 11, §163, the following equation is obtained from which IT can be computed: I = tan-i l = l-i + i-|+- • •• This series converges very slowly. To obtain a more rapidly con- verging series make use of the relation tan~^ 1 = tan"^ ^ + tan~^ |. Then 4 ' (3) (23) + (5) (2^) (7) (2^)^ ^' (3)(33) ^ (5)(3^) (7)(3^)^ 168. Relation between the Exponential and Circular Functions. If it be admitted that the Maclaurin's series expansion e' = l+2 + |2 + ^ + - • •, (1) which was proved for real values of z, is also true when z is imagin- ary, we obtain, on setting z = ix, 6.^ = l-hta;+^-h-|3--|-^-h-|5- + ^ + -|7-+- ■ • ^ , , x^ ix^ , X* , ix^ x^ ix'' , = l+tx-j2- || + |4+^-j6--|7 + - • • (2) On separating real and imaginary parts this becomes e"-l-^ + |4-|g-|---- + 1(2 13 ' 15 17 21 + %-%+■■). (3) 322 CALCULUS [§168 Since (Exercise 1 and Illustration 1, §163), and cosx = l-j2- + |^-ig + x^ , x^ x* . sinx=x-j3+^-i^ + it follows that e« = cos X + i sin X. On changing the sign of x it results that e-»^ = cos X — 1 sin X. Solving equations (4) and (5) for cos x and sin x, and cosx = sinx = 2 e** — e- (4) (5) (6) (7) These interesting relations between the circular and exponential functions are of very great importance. Fig. 122. Fig. 123. If Q represent the vectorial angle in the complex number plane, then it is clear from Fig. 122 that e*' represents a point on the unit circle (circle of radius 1 about the origin as center) in this plane. Further, any complex number a + hi can be put in the form pe". For (Fig. 123) a -^-hi = p (cos + 1 sin Q) = pe*", where p = y/a^ + h"^. 5. e*"-. 7. e^*'. 6. e-'"". 8. 3tV 5e 4 . §169] TAYLOR'S AND MACLAURIN'S THEOREMS 323 Exercises Represent by a point in the complex plane : IX IT 1. 3e^ 3. e^. 2. 2e" 3 , 4. e^. 9. Express the numbers of Exercises 1-8 in the form a + hi. 169. DeMoivre's Theorem. The interesting and important theorem, known as DeMoivre's Theorem, (cos + i sin 0)" = cos n0 + i sin n^ (1) can be easily established by the use of the relation (4) of §168. For, (cos d + i sin 0)» = {e'^Y = e*"^ = cos nd + i sin nd. Exercises Find, by the use of (1), 1. The cube of 1 + i. 1 + iy/^ 4. Ihe cube of 2. The square of ^ • - n^i i -. — 1 — iy/Z 2 5. The cube of ^ -• 3. The cube of ^ 6. The cube of ^ •• In (1), 11 may be a fraction as well as an integer. It will then indicate a root instead of a power. In this case we do not have simply one root: — fl ft (cos d -\- i sin 0)'" = cos [- i sin— > mm (n having been placed equal to — , where m is an integer) but m — 1 additional roots. This follows from the fact that giO _ gi(0 + 2p») (2) where p = 0, 1, 2, 3, 4, • • ■ ,m,m -\- \, • • • . Hence we can write (cos d + I sin ^)™ = [e^"]'" = [e'^" + ^p^^^^ or 1^ t(0 + 2pw) (cos ^ + t sin d)"" = e »» , (p = 0, 1, 2, • • •) ^^' 324 CALCULUS [§170 It would appear at first sight as if there were infinitely many roots corresponding to the infinitely many values of p. But a little consideration shows that when p y m, the roots already found by letting p take the values 0, 1, 2, • • • , w — 1, repeat themselves, since e^" = 1. There are then exactly m mM^ roots of e*^ = cos + I sin d, e- —^ = cos — ' — — + t sm — ^ — —, (4) where p = 0,1,2, • • • , w — 1. Illustration. Find the three cube roots of — 1. (-l)i = (e-)i ^ ^gtu + 2px)j} (p = 0,1,2) t (ir + 2px) = e 3 (p = 0, 1, 2) tV 5jt = e^ , e*', and c ^ . Exercises ie 1. Show that the three cube roots of a + 6t = pe'* are : -y^ « ^i «(g + 2t) i(g -f 4t) ■^^ e 3 J and -^^ c 3 . How would these roots be deter- mined graphically? 2. Find the two square roots of 1 + i. 3. Find graphically the two square roots of l. 4. Find graphically the three cube roots of 1. 170. Indeterminate Forms. It has already been shown that ^ has no meaning. See §25. Thus x-2 has no meaning at a; = 2. Its value at x = 2 is defined as lim ^^ ~ '^ = 4 x=2 X — 2 Similarly sin a tan a has no meaning at a = 0. Its value at a = is defined as lim sin a ^ ^ °=o tana §170] TAYLOR'S AND MACLAURIN'S THEOREMS 325 In general, if </)(a) = and /(a) = 0, the value of the function lim ^xji x = a f(^x) ' -77-v Sii X = a IS denned as The calculation of this limit is simplified in many cases by the application of the law of the mean. See §160. Thus, let it be given that 4>{a) =0 and /(a) = and let 4)(x) and f(x) satisfy the conditions imposed in the statement of the law of the mean. Then lim 0(a;) ^ Hm </>(») + (x - a) 4>' [a + 0\{x - a)] x^a /(^) x^a ji^a) + {x-a) f [a + d^ix - a)] __ lim 4>'[a + ei(x- a)] 0^(a) ^ .^i . . - z^a f'[a + e,{x-a)] /'(a) ' "^ ^6,^ ^^ If (f>'(a) and /'(a) are also zero, we make use of the extended law of the mean. Thus 0(a) + (x - a) 0'(a) + ^^-^V"[a + ^1(0;- a)l lim 4>{x) ^ lim |i •'^ ^ /(«) + (x- a) f'(a) + ^-^—^ria + d^ix - a)] _ lim </> '[a + gi (x - g)] <i>"(a) x^a j"[a + d,{x - a)\ f{a) ' The process is to be continued further if f"(a) and <^"(a) are both zero. Illustration 1. lim e' -1 ^ x=o X Illustration 2. lim e^ — e~' — 2x a:=&0 X — sm a: Hm e- z=0 1 = 1. lim € . + e-x_ 2 ' x=0 1 — cos X lim e^ . _ g-x x=0 sin X lim e' + e-* 1=0 p.na X 2. 326 CALCULUS [§170 The Form — . The same process is employed in evaluating 00 the indeterminate form — -. The proof is omitted. Illustration 3. lim ^ _ lim 2x _ lim 2^ _« X— 00 nx X— oo gx X= 00 ox ' The Form <» . The indeterminate form «> can be thrown 00 into either of the forms - or — ■. Thus 00 lim 3. „„. y _ lim _^ lim 1 _ . =^=0 -^ ^"^ ^ ^=0 tan X ^^0 sec2 x~ Other indeterminate forms are : °o — «> , 1 °°, 0", °°°. Thus, if <p(x) and /(x) become infinite for x == a, (/>(a) — /(a) is defined as ^^^JHx) - fix)]. This expression can be written 1 1 lim r , / V ,/ ^^ _ lim fix) <t>{x) W)f(x) an indeterminate form of the type ^. If (l>{x) becomes infinite and/(x) becomes 1 for x = a, [/(a)]*^"^ is defined as This limit can be calculated as follows. Let y = [/(x)]*^^'. Then logy = 4){x) log/(x) = ^ > W) an indeterminate form of the type -. If lim log /(a;) W) §170] TAYLOR'S AND MACLAURIN'S THEOREMS 327 is found to be c, then lim „ = g« x = a The two remaining forms are evaluated in a manner similar to the last. Many indeterminate forms can be evaluated directly by simple algebraic transformations. Exercises Evaluate the following: lim loga^ 13 lim _^. J- x=l X - i • 2=°' log a; lim 1 -cosg 14. lim ^ lim c» 9=0 cosflsin^fl « lim a; COS a; — sin x 16. ^^ „ ^ X = X 4. lim tan x - sin x 16. ^™ e« tan -• z=0 a; — sin X lim a:" - 1 ^- a;=l X - 1 ^„ lim lim sin 33; • 1=0 sin 2a; 17 lim r_^ L_" ^'' X=l Lx2 - 1 X-l_ i™ [(|-9)ta„; 2 - i=u smzx r r 1 11 ^ lim tan 3a; 19. ^^^ |^j^ - a._ij' g Um ^^ 20.1;- (cos xfot^. x=o sin X — a; 1 lim x^ - X - 6 21. iTod-^)^- ^- x=3 x2 — 9 i:,„ ,„,ta 311+5 22.'-(-x).»«. • I* . 4i! + 1 jj_ hm (3j„ ^)t.n . lim log X 12. li™ '^- „. lim tan fl. CHAPTER XVIII TOTAL DERIVATIVE. EXACT DIFFERENTIAL 171. The Total Derivative. Let z = f{x, y) and let x and y be functions of a third variable t, the time for example. We seek an dz expression for -rr, the derivative of z with respect to t, in terms of dx J dy As an illustration of what is meant, let z denote the area of a rectangle whose sides x and y are functions of t, and at a given instant let each side be changing at a certain rate. The rate at which the area is changing is sought. Returning to the general problem let t take on an increment At. Then x takes on the increment Ax and y the increment Ay, and consequently z the increment Az. We then have 2 = fix, y) (1) 2 -f Az = /(x + Ax, 2/ + Ay) Az=J{z + Ax,y + Ay) - /(x, y) (2) Az = /(x+Ax, y+Ay)-J{x, y+Ay)+f{x, y+Ay)-f(x, y) (3) . Az ^ fix -\-Ax,y + Ay) - /(x, y + Ay) Ax At Ax At fix, y + Ay) - fix, y) Ay "•" Ay At ^ ' Taking the limits of both sides of (4) as At approaches zero, we have dz _ dfjx, y) dx dfjx, y ) dy , . dt dx dt "^ dy dt' ^ ' since Ax and Ay each approach zero as At approaches zero. Equation (5) can be written in the form dz _ 3z dx 3z dy , . dt ~ ax dt "•" ay dt * ^^ 328 §171] TOTAL DERIVATIVE. EXACT DIFFERENTIAL 329 This states that the rate of change of z with respect to t is equal to the rate of change of z with respect to x, times the rate of change of X with respect to f, plus the rate of change of z with respect to y, times the rate of change of y with respect to i. \il = Xy (6) becomes dz dx dz dz dy dx dy dx This formula applies when z = f{x, y) and ?/ is a function of x, e.g., y = <i>{x). Multiplying (6) by dt we obtain dz ^r dx + ^- dy. dx dy (7) This defines dz, which is called the total differential of z. z E I S c F R A f K % 7 Y dy r / Fig. 124. We shall now give a geometrical interpretation of dz. Let P, Fig. 124, be the point {x, y, z) on the surface z = fix, y). Let PC = dy and PA = dx. Then Q is the point {x + dx, y + dy, z + Az). Let PDEF be the plane tangent to the surface at the point P. Then PF is tangent to the arc PR, and PD is tangent to the arc PS. 330 CALCULUS [§171 From F draw FK parallel to AB meeting BE in K. BE = BK-\- KE BK = AF = ^ dx. ox Since FK = PC and PD = FE, triangle KFE is equal to the triangle CPD, and KE ^ CD = ^ dy. dy ^ Therefore ^^ = dx^ + 3y^y' Hence BE = dz. Consequently dz may be interpreted as the increment measured to the tangent plane to 2 = f{x, y) at the point P (x, y, z) when x and y are given the increments dx and dy respectively. Illustration 1. \i z = xy, the area of a rectangle of sides x and y, we obtain by using (7), dz = y dx -\- X dy. The first term on the right-hand side represents the area of the strip BEFC, Fig. 125. The second term the area of DCGH. The difference between Az and dz is the ^ area of the rectangle CFLG, which becomes relatively smaller, the smaller dx and dy become. The above expression could have been obtained by the formula for the B dx E differential of the product of two Fig. 125. variables. Illustration 2. The base of a rectangular piece of brass is 15 feet and its altitude is 10 feet. If the base is increasing in length at the rate of 0.03 foot per hour and the altitude at the rate of 0.02 foot per hour, at what rate is the area changing? Let X denote the base, y the altitude, and z the area. Then z = xy §171] TOTAL DERIVATIVE. EXACT DIFFERENTIAL 331 and dz _ dx dy dt~'^ dt '^^ dt = (10) (0.03) + (15) (0.02). X Illustration 3. z = -• y ^ _ 1, dx y dz dy X ~ y'- and, by (7), dz = — dx 5 dy, y y^ ^' or , y dx — x dy dz = 5 > yi a result which could have been obtained by differentiating the X quotient - by the usual rule. Exercises Find by formula (7) the total differential of each of the following functions : 1. z = x^y. -=^.- 7. 2 = x'e". 2. 2 = XJ/^ 6. 2 = z log y. S. z = e* sin y. a;2 3. 2 = — y 6. z = e* cos y. 9. 2 = e"" cos nx. Find ^f if: dt 10. z = X* cos y- 11. 2 = e* sin y. 12. The radius of the base of a right circular cylinder is 8 inches and its altitude is 25 inches. If the radius of the base is increasing at the rate of 0.2 inch per hour and its altitude at the rate of 0.6 inch per hour, at what rate is the volume increasing? 13. Given the formula connecting the pressure, volume, and tem- perature of a perfect gas, pv = Rt, R being a constant. 11 t = 523°, p = 1500 pounds per square foot, and v = 21.2 cubic feet, find the approximate change in p when I changes to 525° and v to 21.4 cubic feet. 14. If with the data of Exercise 13, the temperature is changing at 332 CALCULUS [§172 the rate of 1° per second, while the volume is changing at the rate of 0.4 cubic foot per second, at what rate is the pressure changing? 15. The edges of a rectangular parallelopiped are 6, 8, and 10 feet. They are increasing at the rate of 0.02 foot per second, 0.03 foot per second and 0.04 foot per second, respectively. At what rate is the volume increasing? 172. Exact Differential. An expression of the form M dx -{- N dy, where M and N are functions of x and y, may or may not be the differential of some function of x and y. If it is, it is called an exact differential. Thus sin y dx -\- X cos y dy (1) is an exact differential, for it is the differential of z = x sin y. The coefficient of dx is ^ = sin y, and that of dy is \- = x cos y. x^ sin y dx -\- x cos y dy (2) is not an exact differential. It is fairly evident from (1) that we dz cannot find a function z = f{x, y) such that v- = x^ sin y and dz -^ — X cos y. dy ^ We seek to find a test for determining whether or not an ex- pression of the form Mdx + N dy (3) is an exact differential. If (3) is the exact differential of a func- tion z, we must have, | = M (4) and 1 = ^. (5) since §172] TOTAL DERIVATIVE. EXACT DIFFERENTIAL 333 Differentiate (4) with respect to y and (5) with respect to x and obtain AU AM (7) (8) dh dM and dy dx ~~ dy d^z dN Since, in general, dxdy " dx d^z dh dydx ~ dxdy it follows that if (3) is an exact differential, we must have dy dx The condition (9) must be satisfied if (3) is an exact differential. It does not follow, however, without further proof, that (3) is an exact differential if (9) is satisfied. It can, however, be shown that this is the case. The proof will be omitted. (3) cannot be an exact differential unless (9) is satisfied and is an exact differ- ential if (9) is satisfied. When an expression of the form (3) is given, the first step is to determine whether or not it is an exact differential by applying the test (9). If it is an exact differential, the next step is to find the function z of which it is the differential. This step will be illustrated by integrating several differentials for which the functions from which they were obtained by differentiation are known. Illustration 1. If z = a;^ + 2xhj -\- y^ -\- C, ^' = dic'^'' + dl^y = {3x^ + 4xy)dx + (2x^ + 2y)dy. If then we are given the exact differential dz = (3a;2 + ^xy)dx + {2x^ + 2y)dy and arc required to find the function of which it is the differential, we note first that dz Then d^x = ^^' + ^^2/. 2 = a;3 _|_ 2xhj + a Junction of y alone. 334 CALCULUS [§172 And this function of ?/ is to be so determined that dz Clearly the term 2x^ is obtained by taking the derivative with respect to t/ of 2x^2/, a term already found, and consequently it is not to be added. 2y is the derivative of y^. y^ is then the function of y which is to be added to the terms already found. Further an arbitrary constant is to be added since its differential will be zero. Then z = x' + 2xh/ + y^ + C is the function whose differential was given. If, as is usually the case, it had been given that (3x* + ^xy)dx + (2x2 + 2y)dy = (10) it would have been required to find a function of x and y such that its differential would be zero. Now the first member is, as we have seen, the differential of z = x^ + 2xh) + y\ But, if (fe = 0, 2 = C. Then x» + 2xhi + y^ = C is the relation between x and y which satisfies the given equation. Illustration 2. If z = e* cos y -{- x^ -\- sin y + y^, dz = (e* cos y -\- 2x) (fx + ( — e* sin y + cos y + ^y^) dy Now let it be given that (e* cos y + 2x) dx -\- (— e' sin y -\- cos y + Sy^) dy = 0. (11) From its derivation we know that the left-hand member is an exact differential, dz. Let us proceed to find z as if it were unknown. dz Then -— = e" cos y + 2x. z = e* cos y + x^ + a function of y alone. (12) §173] TOTAL DERIVATIVE. EXACT DIFFERENTIAL 335 The function of y is to be so determined that dz ^ = - e' sm y + cos rj -{■ dy'^. (13) The first term is evidently obtained by differentiating e' cos y, a term already found in (12). The remaining two terms in (13) are obtained by differentiating sin y + y^. These are to be added to the terms already found in (12). Then z = e' cos y -{- x^ -\- sin y + y^. But, since dz = 0, z = C. Hence e" cos y -{- x^ -{- sin y -\- y^ = C is a solution of (2). Illustration 3, Integrate if possible the equation {e' y+sin y-\-2x) da;+(e'+x cos y-\-e''-\-2y—sin y) dy = 0. (14) We have first to determine whether or not the first member is an exact differential. Apply the test (9), dM ~ = e' + cos y. dN —— = e* + cos y. Hence (9) is satisfied and the first member of (14) is an exact differential. On integrating the coefficient of dx with respect to X we obtain c^y -{- xsiny -{- x"^. To this we have to add gV ^ y2 _|_ pQg y^ the terms which arise from the integration of the coefficient of dy and which contain y alone. (The other terms in the coefficient of dy arise from the differentiation of terms already found by integrat- ing the coefficient of dx.) Then the solution of (14) is e^y + xsiny + x^ + e" + y^ + cosy = C. 173. Exact Differential Equations. Equations involving differ- entials or derivatives are called differential equations. Those of the type Mdx + N dy = (1) 336 CALCULUS [§174 where the first member is an exact differential, are called exact differential equations. The equations (10), (11), and (14) of Illustrations 1, 2, and 3, §172, are exact differential equations. The process of finding the relation between y and x, which when differentiated gives a certain differential equation, is called the integration of the equation. The procedure in dealing with an equation of type (1) is to determine first whether or not it is exact by applying the test (9), §172. If it is, integrate the coefficient of dx with respect to x and to this result add those terms which contain y only, which are obtained by integrating the coefficient of dy with respect to y. Exercises Are the following differential equations exact? Integrate those which are exact. 1. Sx'^y^dx +2xhjdy = 0. 2. — cos(-)dx cos (-) dy = 0. y \yl y' \y/ 3. y e'" (1 + X + y) dx + X e'" {I + x + y) dy = 0. 4. y e'" dx + X e'" dy = 0. 5. {x^y + 2x) dx - {Sx^y - 5x) dy = 0. 6. (p+l)dx- [^ + 2y)dy=0. 7. e-y(2 + -^^ - j^ e^2 + ^^dy=0. 174. In §155 the envelope of a family of curves was defined, and its parametric equations were found to be fix, y, c) =0 (1) We shall now show that the envelope is tangent to each curve of the family of curves (1). At a given point {x, y) of the curve determined by giving c a particular value in (1), the slope of the tangent is found from the equation dj.dfdj,^^ dx dy dx ' §174] TOTAL DERIVATIVE. EXACT DIFFERENTIAL 337 If the point also lies upon the envelope its coordinates satisfy (1) and (2). The equation of the envelope can be regarded as given by (1) where c is the function of x and y found by solving (2) for c. On differentiating (1) with respect to x, regarding c as a function of x and y, the slope, -r-, of the tangent to the en- velope is given by dx dy dx dc dx ' dc _ dc dc dy dx dx dy dx where fif But on the envelope ^ = 0. Hence (4) becomes ^ + ^ ^ = (5) dx dy dx ' Equations (3) and (5) show that the slope of the tangent line to the envelope at the point (x, y) is the same as the slope of the tan- gent line at the same point to a curve of the family (1). Hence the envelope is tangent to each curve of the family of curves (1). 22 CHAPTER XIX DIFFERENTIAL EQUATIONS 175. Differential Equations. An equation containing deriva- tives or differentials is called a differential equation. If no deriva- tive higher than the first appears it is called a differential equation of the first order. If the equation contains the second, but no higher derivative, the equation is said to be of the second order. And so on. Numerous differential equations have already occurred in this course. We shall now consider the solution of differential equations somewhat systematically. 176. General Solution. Particular Integral. Let fix, y,c)=0 (1) be any equation between x, y, and the constant c. If (1) is differentiated with respect to x there results the equation Fix, y, y', c) = 0. (2) Between (1) and (2) the constant c can be eliminated giving the differential equation of the first order <l>{x, y, y') = 0. (3) Equation (3) follows for any value of the constant c. Let /(x, y, ci, C2) = (4) be an equation involving two constants, Ci and Co. By differ- entiating (4) we obtain Fix, y, y', ci, d) = (5) and <t>ix, y, y', y", ci, c^) = 0. (6) Between equations (4), (5), and (6), Ci and C2 may be eliminated giving the differential equation of the second order Hx, y, y', y") = 0. (7) 338 §178] DIFFERENTIAL EQUATIONS 339 From the equation (1) containing one arbitrary constant the differential equation of the first order (3) is obtained. From the equation (4) containing two arbitrary constants the differential equation of the second order (7) is obtained. In like manner from a relation between x and y containing n arbitrary constants a differential equation of the n*^ order is obtained by differentiat- ing, and eliminating the constants. Equation (1) is a solution of equation (3). It is called the general solution and involves one arbitrary constant of integration, c. Equation (4) is called the general solution of (7). It involves two arbitrary constants, or constants of integration. It can be shown that the general solution, or general integral, of a differential equation contains a number of arbitrary constants, or constants of integration, equal to the order of the differential equation. A particular integral is obtained from the general integral by giving particular values to the constants of integration. 177. Exact Differential Equations. This type of differential equation was discussed in §173. 178. Differential Equations; Variables Separable. The vari- ables X and y are said to be separable in a differential equation which can be put in the form f(x) dx + </>(?/) dy = 0. The first member is equal to a function of x alone multiplied by dx plus a function of y alone multiplied by dy. Illustration 1. (1 + y'^)x dx-\- {l-\- x^)y dy = 0. On dividing by (1 + ?/^)(l + x^) this equation becomes xdx y dy _ l + x^ ' 14-2/ Integration gives ilog (1 + x^) + ilog (1 + 2/2) = C. This reduces to '1> (1 + x^){\ + 2/2) = e2C = c. or 340 CALCULUS I§179 Illustration 2. Vl - y^ dx + \/l - x2 rf?/ = 0. Then dx dy ^ ^ Vl -ic2 Vl - 2/' ' and the variables are separated. Integration gives sin~i X + sin~i y = C. Take the sine of each member, observing that the first member is the sum of two angles, and obtain a;Vl - 2/' + yVi - x"" = sin C = d. Exercises Solve the following differential equations: 1. (1 - x)dy - (1 + y)dx = 0. Ans. (1 + y){l - x) = C. 2. sin X cos y dx = cos x sin y dy. 8. (x - Vl + x^) Vl + 2/^ dx = (1 + x^)dy. 4. ^^ = 5y^x. f. dy y" + 4?/ 4- 5 _ **• dx "^ x2 + 4x + 5 ~ 6. (1 + a:)dj/ = y(l — i/)dx. ^ns. ?/ = c(l + x)(l — y). 7. (1 - x)ydx + (1 + j/)x dy = 0. dv 8. ^ + e'2/ = e'y«. 9. (x» + j/x*)di/ - (t/2 - x?/2)dx = 0. dy , ^ dy 11. 3e* sin y dx + (1 — e') cos y dy = 0. 12. (xy + x»2/)dy - (1 + y^)dx = 0. Ans. (1 4- x')(l + y') = ex*. 179. Homogeneous Differential Equations. The differential equation Mdx-\-N dy = Q (1) is said to be homogeneous if M and iV are homogeneous functions of X and y of the same degree. §179] DIFFERENTIAL EQUATIONS 341 A function f{x, y) of the variables x and y is said to be homogen- eous of degree n if after the substitutions x = Xx', y = \y' have been made, fix, y) = X-/(x', y'). Thus ax^ + bxy -\- cy^ is liomogeneous of degree 2. For, on making the substitutions indicated, it becomes \\ax'^ + bx'y' + cy'^). The expression ax^Vx^ + 1/2 + 6x3 tan-» (-\ is homogeneous of degree 3. For, after the substitutions indi- cated above, it becomes X3 fox'^^/x'^ + y'^ + 6x'3 tan-i Z^;) ] • A homogeneous differential equation of the form (1) is solved by placing y = vx, and thus obtaining a new differential equation in which the variables, v and x, are separable. Illustration: (x2 + 2/2) dx + 3xy dy = 0. Let y = vx. Then dy = V dx -\r X dv, and x2(l + v^) dx + Svx%v dx -\-x dv) = 0. x2(I + 4«;2) dx + 3^3 dv = 0. Separating the variables dx Svdv _ T "•" 1 + 4i;2 " "• log[x(l +4i;2)i] = C, x(l 4-4i'2)i = Ci. 342 CALCULUS [§180 y On substituting t; = - we obtain as the solution of the given equation or x\x^ + 4?/2)3 = Ci. Exercises Solve the following differential equations ^ _ ^„ #. dx ~ ^y dx 2. x'^y dx - (x' + 2/3) di/ = 0. 3. (8y + 10a;) dx + (5y + 7x) dy = 0. 4. (2\/x?/ — x) dy + 2/ dx = 0. ?/ ^2/ 2/ 6. X cos — -J- = y cos X. X dx X dy 7. x ^ - y = Vx'' - y\ 8. {y - x) dy + ydx = 0. 180. Linear Differential Equations of the First Order. The equation where P and Q are functions of x only, is called a linear differential equation. It is of the first degree in y and its derivative. Multi- ply the equation by ' Pdx and obtain ./' iPdxrdy , „ 1 {pdx ^ (2) The left-hand member is the derivative of fpdx eJ y, as may be confirmed by differentiating this product. The inte- gration of (2) gives }180] DIFFERENTIAL EQUATIONS 343 e* f'-"v = JQeJ'""dx + C. Illustration 1. t+^y-"'- Here P = x and Q = x\ Then I Pdx \ xdx T X- Multiply both members of (3) by c 2 . Integration gives a;2 /. a;2 a;2 a;2 = e^x^ - 2e2 + C. Hence a;2 J/ =x2-2 + Ce~"^. Illustration 2. I Pdx (^ oJ = P. '' = Multiply both members of (4) by x. Integration gives xy or = ^ + x^* + 2x2 + C, (3) ^ + l^=:.2 + 3^+4. (4) dx x*^ x' C* y = ^ + x24-2x + -- This illustration is inserted to call attention to the well-known simple relation e'"«^ = x, which there will be frequent occasion 344 CALCULUS [§181 CO use in solving equations of this type. It should be recalled that e''^°«^ = e'°«(^"> = x". Thus e — loga; _ t X Exercises 1. dy dx + 2xy = e ■^\ 2. dy dx + y cos X = sin 2x. . dy , 3. cos'^x-T — r y = tan x. 4. (x« + l)^ + 2xy = 4x2. '•g + ^ = (- + «•• 6. x(l - x«) dy + (2x2 _ i)^ ^^ = ^x^ ^^.^ „ dy y 1. -/ -n- = e'x". dx X 8. (1 + x^) dy ^- (xy -^ dx = Q. ^ dy ,1 -2x 10. (1 + y"^) dx = (tan-iy - x) dy. 181. Extended Form of the Linear Differential Equation. An equation of the form % + Pv = Qr (1) is easily reduced to the linear form. For, on dividing (1) by y, we obtain The first term of the left-hand member of (2) is, apart from a constant factor, the derivative of ?/""+', which occurs in the second term. If we let z = 2/~"^^ we obtain the linear differential equation or g _ (n - l)Pz = -in- 1)Q. §1821 DIFFERENTIAL EQUATIONS 345 Illustration 1. dv T — h 2/ cos X = y* sin 2x. Dividing by y* Let 2 = 7/-3. Then .dy,, . „ y~* -j~ + y~^ cos X = sin 2x, ,-4 ^ = _ 1 ^ and the equation becomes dz dx ^ dz , . „ — 5 ;i — \- z cos X = sin 2x, or 3 Zz COS X = — 3 sin 2a;. dx This equation can be readily solved by §180; y~^ is to be sub- stituted for z in the result. Exercises 1- ^- + ~ 2/ = a;V. 4. (1 - x2) ^ - xy = axy^. dy 2. ;^^ + 2/ = xy3. ^'t+ly = '^'y^- 3.3,^1-7,3 = X + 1. 6. X ^ + , = ,2 log X. 7 '^^ + 2 , - ^'. 182. Applications. Let there be an electric circuit, whose resistance is R, whose coefficient of self-induction is L, and which contains an electromotive force, which at first we shall suppose constant and equal to E. It is required to find the current i at any time t after the time < = 0, at which the circuit was closed. The equation connecting the quantities involved is readily set up. The applied E.M.F., E, must overcome the resistance of the circuit and its self-induction. The former requires an E.M.F. equal to iR, and the latter an E.M.F. proportional to the time rate of dz dz change of current, viz., j, and equal to L ^. The applied E.M.F., E, must equal the sum of these two E.M.F.'s. 346 CALCULUS [§182 Hence ^ di , ^. ^ ,,. Lj^ + Ri = E, (1) The student will show that, if i equals zero when t equals zero, the solution of this linear equation is Er ^h If the battery or other source of E.M.F. is suddenly cut out of the circuit, the current falls off in such a way that the differential equation Lj^ + Ri = (3) is satisfied. Show that the law at which the current falls off is t = ioe-L^'- '»>, (4) if the instant at which the battery is cut out is the time 1 = 1^ and if the current at this instant is i = Iq. If the E.M.F. is variable, the relation between the quantities involved in the circuit is still governed by (1), Lj^ + Ri^E, (1) in which E is now variable. Suppose E = Eo sin cof. This sup- poses that an alternating E.M.F. is acting in the circuit. The differential equation to be solved is di L-r. + Ri = Eo sin cat. (5) Show that *« ~ L R , R^ + co^L y- sin cat — 03 cos coHe ^ + C ft (R sin cat — caL cos cat)e + C Eq . , . , X ^ ^f V<R' + w^L^ =r-rr- siu (cof — 0)e + C, §183] DIFFERENTIAL EQUATIONS 347 where sm <f) = Then, cos d> = — ,- =. R Jp — ^ t i = ^ ° sin (co< - 0) + Ce ^ Since the last term becomes negligible after a short time because of the factor e ^ > it is scarcely necessary to determine C. On droppmg out the last term as unimportant except in the immediate vicinity oit =0, we have The current, therefore, alternates with the same frequency as the E.M.F., but lags behind it and differs from it in phase by 4>- It is to be noted that the maximum value of the current is not -^5- K JP but . • The quantity -s/R^ + oj^L^ replaces, in alter- V /2^ + co^L^ nating currents, the resistance R of the ordinary circuit. It is called the impedance of the circuit. 183. Linear Differential Equations of Higher Order with Con- stant CoeflScients and Second Member Zero. A typical differen- tial equation of this class is the following: d"y , d^-^y , d"~hj , , dy , ^ /,s where Oq, Oi, • • •, a„ are constants. As the equations of this class which occur in the applications are usually of the second order we shall confine our discussion in this article to linear differential equations of the second order. Consider 348 CALCULUS [§183 Let us assume that 2/ = e-"* (3) and find, if possible, the values of m for which (3) is a solution of (2). The substitution of (3) in (2) gives e"" {aom^ + aim + a^ = 0. (4) The first factor cannot vanish. The second, equated to zero, gives a quadratic equation in m. Call its roots nii and mt. Then (3) is a solution of (2) if m has either of the values Wi or mo, the roots of OqW^ + aim + a2 = 0. (5) The equation (5) in m, obtained from the given differential equa- d'^v du tion by writing m^ for y-^ and to for -v- is called the auxiliary equation. Two solutions of (2) are y = gmii and y = e'^i*. Furthermore, y = Cie-"!" is a solution of (2). For, after the substitution of this value of y in (2), Ci can be taken out as a common factor and the other factor vanishes in accordance with (4) or (5). In the same way, y = de'^i' is a solution of (2). And finally the sum of the two solutions y — Cie""!^ + C2e"'2* (6) is a solution of (2). This can be seen by substituting in (2) and recalling that nii and mo are roots of (5). When mi is not equal to m2, (6) is known as the general solution of the differential equa- tion (2). It contains two arbitrary constants, the number which the general solution of a differential equation of the second order must contain. The values of these constants are determined in a particular problem by two suitable conditions. Illustration. §184] DIFFERENTIAL EQUATIONS 349 The auxiliary equation is m^ — bm + 6 = 0, [m - 2){m - 3) = 0. Hence mi = 2, mz = 3. The general solution is then y = Cie^' + C26''. Exercises »-g-l-''-°- 2.g-4, = 0. »-g-4^'»-»- *-g + |->^-»- '■2+^t-o- 184. Auxiliary Equation with Equal Roots. The method just given fails when the auxiliary equation has equal roots, mi = W2. For, equation (6), §183, becomes y = de""!' + C2e'"2* = (Ci + C2)e-^'. But Ci + C2 is an arbitrary constant and the solution contains only one arbitrary constant instead of two. When the auxiliary equation has equal roots, mi = 7^2, equation (2) can be written in the form Its general solution is This solution can be verified by direct substitution. Illustration. ^_4^+4v = 0. dx^ dx^ y "• The auxiliary equation is m^ — 4m + 4 = 0. TMi = m2 = 2. 350 CALCULUS [§185 The general solution is 2/ = (C, + C^x)e^'. Exercises l-^^ + '^g + o-"- 185. Auxiliary Equation with Complex Roots. If the auxiliary equation has complex roots the general solution can be written in a form different from (6), §183. The importance of the result will be evident at once when it is observed that it contains the harmonic functions sine and cosine. If the coefficients of the given differential equation (2), §183 are real, and if mi and niz are complex, they must be conjugate imaginary numbers. Let THi = a -\- ih. Then mz = a — ih. Then (6) becomes y = Cie" + '*' + Cze "' ~ *'" Now, by (4) and (5), §167, gift* = cos bx + i sin bx Q-ibx — COS bx — i sin bx. Then y = c* [(Ci + C-i) cos bx + i{Ci — Cz) sin 6a;] On placing Ci + d = A and i{Ci — Cz) = B, we obtain y = go* (^ cos bx -\- B sin bx) = e"' C cos (6a; — <f>). In the last form the two arbitrary constants of integration are C and <f). Illustration 1. The auxiliary equation is m2 + 4m + 13 = 0. §186] DIFFERENTIAL EQUATIONS 351 I Hence m = -2 ± Si. Then y = e-2x (J^ COS Sx -\- B sin 3a;). Illustration 2. 2^ + ^y = o. Whence Then m^ + 4 = 0. m = ± 2i = + 2i. T/ = A cos 2x + i5 sin 2x. Exercises d^ d^9 5^ + .^. = o. 2.^, + 9y = 0. dx^+'^ = 0- 186. Damped Harmonic Motion. The resistance offered by the air to the motion of a body through it, is roughly proportional to the velocity, if the velocity is a moderate one. In §81, the differ- ential equation of the motion of the simple pendulum was derived on the assumption that the force of gravity was the only force acting upon the bob of the pendulum. If the resistance of the air is also taken into account we shall have to add to the second member of the equation, I -xr^ = — ^ sin 0, a term, — 2kl -r. pro- portional to the velocity I -rr- See equation (1), §81. The differ- ential equation of the motion is then idW . ^ ^,^dd ... l^=-g.me-2kl-^. (1) The negative sign is used before the last term because the force due to the resistance of the air acts in a direction opposite to that 352 CALCULUS I §186 of the motion. The advantage of choosing 2k as the proportion- ality factor instead of k will appear later. A; is a positive constant. From (1) we obtain j;|+2.f + f.„. = 0. (2) As in §81 assume that 6 is small and replace sin 6 by 9. Also let g I d^e dd This is a linear differential equation of the second order with constant coefficients and can be solved by the method of §185. The auxiliary equation is m^ + 2km + w^ = 0, whence m = - k ± \/k'^ - o) 2. When the velocity is not very great, as in the case of an ordinary pendulum, k is very small for air and is much less than oj. The expression under the radical sign is negative. We write then m = - k ± i Vw^ - A;2, oj^ — k^ being positive. The solution of (3) is e = Ae-*' cos [t^/(a^ — k'^ - e], or, multiplying both sides by I and replacing Al by B, s = 5e~*' cos[t\/(t)^ — k- — c]. The motion is a damped harmonic motion. The amplitude de- 2t creases with the time. The period . — is a little greater ■yct}^ — k'^ 2x . than — , the period of the free motion. Since k is very small in comparison with w, we can, for an approximate solution of our problem, neglect k^ in comparison with co^. Equation (4) becomes s = 5e-*' cos (aj< — e). (5) This represents the motion with a high degree of approximation. §186] DIFFERENTIAL EQUATIONS 353 The arbitrary constants B and e can be determined by suitable initial conditions. For example, let it be given that s = sq and ds '77 = when t = 0. On differentiating (5) we obtain ds jT= Be-'"[ — k cos (cjt — e) — CO sin(co< — e)]. (6) For f = we obtain from (5) and (6) So = B cos € = B( — k cos e + w sin e). From the latter of these two equations tan € = — . CO From the former 1 -\ — i = to the degree of approximation used above. We have then as the approximate equation of motion s = SqC"*' cos (_ut — e) (7) where k k € = tan~' — = — J approximately. Since k is very small, c is very small. It follows from (5) and (7) that the period of the pendulum in the case just considered is very little different from that of the same pendulum swinging in a vacuum. The amplitude of the swing, however, is affected and diminishes continually with the time. 23 INDEX (Numbers refer to pages) Acceleration, 52, 83, 136, 138 angular, 138 average, 52 Algebraic function, 4, 55 definition of, 56 rational, 6 Alternating series, 316 Angle, between two lines, 321 between two planes, 236 Angular acceleration, 138 velocity, 138 Anti-derivative, 46, 49 Applications, 218 Arc, differential of length of, 100, 181 length of, 218 Area, by double integration, 253, 258 moment of inertia of, 281 polar coordinates, 183, 218 of surface of revolution of, 112, 219, 274 under a curve, 75, 218 Arithmetic mean, 116 Axes, coordinate, 228 Base, change of, 147 naperian, 146 natural, 146 147 199 func- Cable, parabolic, 82 Catenary, 161 Center, of curvature, 290, 294 of gravity, 250, 254, 274 of mass, 263 Centroid, 263, 265, 274, 281 of a line, 268 of an area, 268 • of a solid, 268 Change of base, logarithm, of limits of integration, Circle, curvature of, 290 of curvature, 290 Circular and exponential tions, relation between. 321 Circular functions, 122, 321 Comparison test, 316 Complex numbers, 323 Compound interest law, 155 Computation of IT, 321 of logarithms, 320 Concavity of curve, 90 Conicoid, 242 Constant, 1 Continuous function, 31 Contraction of curve, 7 Convergence of series, 315 Convergent series, 303 Co6rdinate axes, 228 planes, 228 Cosines, direction, 230 Curvature, 289 approximate formula for, 293 center of, 290, 294 356 356 INDEX Curvature, circle of, 290 of a circle, 290 defined, 289 parametric equations, 292 radius of, 290 Curve, contaction of, 7 direction of, 178 elongation of, 7 orthographic projection of, 7 shear of, 7 translation of, 7 Curves of hyperbolic type, 2 maxima and minima points of, 20 of parabolic type, 2 Cylindrical surfaces, 244 Damped harmonic motion, 351 Definite double integrals, 255 integrals, 88, 103, 104 DeMoivre's Theorem, 323 Dependent variable, 1 Derivative, 21 first, 21 of a constant, 43 of a function of a function, 66 of a quotient, 58 of circular functions, 122, 124, 131 of exponential functions, 145 of logarithmic functions, 145 of sin u, 122 of the product of two func- tions, 57 of the sum of a function and a constant, 39 of the sum of a constant and a variable, 56 of the sum of two functions, 43 Derivative of m„, 42, 59 second, 70 total, 328, 329 Derivatives of higher order, par- tial, 247 partial, 246 Differential, 88, 95 exact, 328, 332 of length of arc: polar coordinates, 181, 218 rectangular coordinates, 100, 218 Differential equation, 338 exact, 335 linear, 342 of higher order, 347 order of, 338 variables separable, 339 Diiferentiation, 31, 97 implicit, 44 logarithmic, 153 Direction cosines, 230, 239 of curve, 178 Distance between two points, 230 of a point from a plane, 237 Divergent series, 303 Double integration, 251, 253, 258, 255 Duhamel's theorem, 105 Element of integration, 113 Ellipsoid, 242 Elliptic paraboloid, 244 Elongation of curve, 7 Envelope, 289, 296, 336 of normals, 300 Equation, differential, 338 exact differential, 335 homogeneous differential, 340 INDEX 357 Equation, linear differential, 342, 347 of first degree in x, y and z, 233 of a plane, 233 intercept form of, 234 normal form of, 232 Equations of a line, 238 parametric, 67 Evolute, 289, 294, 300 the envelope of the normals, 300 Exact differential, 328, 332 equation, 335 Exponential functions, 9, 145, 321 Extended law of the mean, 307 Falling body, 22 Family of curves, 297 Formulas, integration, 185, 201, 203, 205, 206, 207 Formulas, Wallis', 204 Fractions, partial, 211 Function, 1 a", 9 a cos X + '^ sin x, 9, 165 a cos X, 9 ax' +px + y, 165 6 sin X, 9 sin X, 9 maximum and minimum val- ues of, 20 mx ± y/a" - x% 168 xS 16 Functions, algebraic, 4, 5, 56 circular, 122, 321 continuous and discontinu- ous, 31 exponential, 145, 321 hyperbolic, 159 implicit, 44 Functions, logarithmic, 145 power, 1 rational, 6 integral, 6 transcendental, 6 transformations of, 10 General solution of a differential equation, 338 Harmonic motion, 140 damped, 351 Homogeneous differential equa- tion, 340 Hyperbolic functions, 159 paraboloid, 244 Hyperboloid, of one sheet, 244 of two sheets, 244 Implicit differentiation, 44 Improper integrals, 218, 223, 226 Increments, 13 Indefinite integrals, 104 Independent variable, 1 Indeterminate forms, 30, 31, 303, 304 Infinite limits of integration, 226 Infinite series, 303 Infinitesimals, 29, 88 limits of ratio of two, 93 order of, 91 Infinity, 29 Inflection, point of, 70 Integral, the I sec' X dx, 204 I e°* sin nx dx, 202 I e "' cos nx dx, 202 Integrals, improper, 218 Integration, 46, 49, 185 358 INDEX Integration, by parts, 201 double, 251, 253, 258 formulas, 185, 201, 203, 205, 206, 207 of expressions containing, g-T " + bx + c,.190 Va^ + x^ V a^ - x2, Vx^ - a^, 196 powers of x and of a + hx, 192 . Ca smx -\- I of I — : J c sin a; + .i bcos x dx, 209 J cos X of powers of trigonometric functions, 193 successive, 250 triple, 250, 260 Intercept form of the equation of a plane, 234 Inverse functions, 67 Involutes, 302 Law of the mean, 306 extended, 307 Length of arc, polar coordinates, 181, 218 rectangular coordinates, 100, 111, 218 Limit, definition of, 27 of the quotient of two infinitesimals, 93 of S/(x)x, 102 Limits, infinite, of integration, 226 of integration, change of, 199 theorems on, 30 Line, direction cosines of, 230 equations of, 238 Linear differential equation, of first order, 342 of higher order, 347 Loci, theorems on, 10 Logarithmic differentiation, 153 Logarithmic functions, 145 Logarithms, computation of, 320 Maclaurin's series, 311 theorem, 303, 310 Maxima and minima, 20, 60, 165 applications of, 174 by limits of curve, 169 determined by derivative, 169 second derivative test for, 172 Maximum and minimum values of functions, 60 Maximum defined, 20 Mean, arithmetic, 116 Mean value of a function, 117 Mean, law of the, 306, 307 Minimum defined, 20 Moment, 263 Moment arm, 263 Moment of inertia, 250, 277 of area, 280, 281, 283 of a solid, 284 polar coordinates, 283 translation of axes, 280 with respect to a plane, 285 Naperian base, 146 Natural base, 146 Normal form of the equation of a plane, 232 Normal, length of, 68 Normals, envelope of, 300 Octant, 229 Order of differential equation, 338 Orthographic projection of curve, 7 INDEX 359 Pappus, theorems of, 274 Parabolic cable, 82 Paraboloid, elliptic, 244 hyperbolic, 244 of revolution, 24 Parallel planes, 236 Parameter, 297 Parametric equations, 67, 292 Partial derivatives, 246 of higher order, 247 Partial fractions, 211 Particular integral of a differen- tial equation, 338, 339 Path of a projectile, 85 Pendulum, the simple, 141 Per cent, rate, 159 Perpendicular planes, 236 Plane, general equation of, 234 intercept form of equation of, 234 normal form of equation of, 232 Planes, angle between two, 236 coordinate, 228 parallel, 236 perpendicular, 236 Point of inflection, 70 Points, distance between, 230 Polar coordinates, 178 area, 183, 258 centroid, 276 differential of arc, 181 direction of curve in, 178 moment of inertia in, 283 Polynomials, 4 Power function, 1, 33 derivative of, 33, 35, 41, 42 hyperbolic type, 2 law of, 3 Power function, parabolic type, 2 Power series, .'505 Projectile, path of, 85 Quadric surface, 242 Quotient, derivative of, 58 Radius of curvature, 290 approximate formula for, 293 Radius of gyration, 278 Rate of change, 37 Rational algebraic function, 6 Relative rate, 159 Rolle's theorem, 305 Second derivative, 70, 172 Series, 303 alternating, 316 convergence of, 315 convergent, 316, 303 divergent, 303, 315 infinite, 303 Maclaurin's, 311 power, 305 Taylor's, 311 test for convergence, 315 Shear of curve, 8 Simple harmonic motion, 140 pendulum, 141 Slope of tangent line, 17 Solid geometry, 228 Solid of revolution, 109, 218 surface of, 219 volume of, 109, 218, 260 Solution of differential equation, 338 Subnormal, length of, 68 Subtangent, length of, 68 Successive integration, 250 Surface of revolution, 112, 219 Surfaces, cylindrical, 244 360 INDEX Surfaces of revolution, 240, 274 quadric, 242 Symmetric form of the equations of a line, 238 Tangent, length of, 68 slope of, 17 Taylor's series, 311 theorem, 303, 308 Tests for convergence, 316 Theorems of Pappus, 274 Total derivative, 328 differential, 329 Transcendental functions, 6 Transformation of functions, 10 Translation of curves, 6 Triple integration, 250, 260 Variable, 1 Velocity, 16, 52, 136, 138 average, 16 of a falling body, 22 Volume by triple integration, 260 of a solid of revolution, 109, 218, 274 Wallis' formula, 204 Water pressure, 114, 219 Work done by a variable force, 77, 107, 219 Date Due QTR inAM oh: 9 77 RBTD „, \h 77 G|U (S PRINTED IN U. 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